{"id":33007,"date":"2022-01-25T17:53:27","date_gmt":"2022-01-25T12:23:27","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=33007"},"modified":"2025-10-23T13:36:23","modified_gmt":"2025-10-23T08:06:23","slug":"important-questions-for-class-10-maths-chapter-5-arithmetic-progressions","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-5\/arithmetic-progressions\/","title":{"rendered":"Important Questions for Class 10 Maths Chapter 5: Arithmetic Progression"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_37 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" style=\"display: none;\"><label for=\"item\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input type=\"checkbox\" id=\"item\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1' style='display:block'><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-5\/arithmetic-progressions\/#Arithmetic_Progressions_Class_10_Important_Questions_Very_Short_Answer_1_Mark\" title=\"Arithmetic Progressions Class 10 Important Questions Very Short Answer (1 Mark)\">Arithmetic Progressions Class 10 Important Questions Very Short Answer (1 Mark)<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-5\/arithmetic-progressions\/#Arithmetic_Progressions_Class_10_Important_Questions_Short_Answer-I_2_Marks\" title=\"Arithmetic Progressions Class 10 Important Questions Short Answer-I (2 Marks)\">Arithmetic Progressions Class 10 Important Questions Short Answer-I (2 Marks)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-5\/arithmetic-progressions\/#Arithmetic_Progressions_Class_10_Important_Questions_Short_Answer_-_II_3_Marks\" title=\"Arithmetic Progressions Class 10 Important Questions Short Answer \u2013 II (3 Marks)\">Arithmetic Progressions Class 10 Important Questions Short Answer \u2013 II (3 Marks)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-10\/maths\/chapter-5\/arithmetic-progressions\/#Arithmetic_Progressions_Class_10_Important_Questions_Long_Answer_4_Marks\" title=\"Arithmetic Progressions Class 10 Important Questions Long Answer (4 Marks)\">Arithmetic Progressions Class 10 Important Questions Long Answer (4 Marks)<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"Arithmetic_Progressions_Class_10_Important_Questions_Very_Short_Answer_1_Mark\"><\/span>Arithmetic Progressions Class 10 Important Questions Very Short Answer (1 Mark)<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1. Find the common difference of the AP \\(\\frac{1}{p}, \\frac{1-p}{p}, \\frac{1-2 p}{p}, \\dots\\). (2013D)<\/strong><br \/>\nSolution: The common difference,<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130854\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-1.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 1\" width=\"188\" height=\"90\" \/><\/p>\n<p><strong>Question 2. Find the common difference of the A.P. \\(\\frac{1}{2 b}, \\frac{1-6 b}{2 b}, \\frac{1-12 b}{2 b}, \\ldots\\). (2013D)<\/strong><br \/>\nSolution: The common difference, d = a<sub>2<\/sub> \u2013 a<sub>1<\/sub> = \\(\\frac{1-6 b}{2 b}-\\frac{1}{2 b}\\)<br \/>\n= \\(\\frac{1-6 b-1}{2 b}=\\frac{-6 b}{2 b}\\) = -3<\/p>\n<p><strong>Question 3. Find the common difference of the A.P. \\(\\frac{1}{3 q}, \\frac{1-6 q}{3 q}, \\frac{1-12 q}{3 q}, \\dots\\). (2013D)<\/strong><br \/>\nSolution: Common difference, d = a<sub>2<\/sub> \u2013 a<sub>1<\/sub> = \\(\\frac{1-6 q}{3 q}-\\frac{1}{3 q}\\)<br \/>\n= \\(\\frac{1-6 q-1}{3 q}=\\frac{-6 q}{3 q}\\) = -2<\/p>\n<p><strong>Question 4. Calculate the common difference of the A.P. \\(\\frac{1}{b}, \\frac{3-b}{3 b}, \\frac{3-2 b}{3 b}, \\dots\\). (2013D)<\/strong><br \/>\nSolution:<br \/>\nCommon difference, d = a<sub>2<\/sub> \u2013 a<sub>1<\/sub>= \\(\\frac{3-b}{3 b}-\\frac{1}{b}\\)<br \/>\n= \\(\\frac{3-b-3}{3 b}=\\frac{-b}{3 b}=\\frac{-1}{3}\\)<\/p>\n<p><strong>Question 5. Calculate the common difference of the A.P. \\(\\frac{1}{3}, \\frac{1-3 b}{3}, \\frac{1-6 b}{3}, \\dots\\) (2013OD)<\/strong><br \/>\nSolution:<br \/>\nCommon difference, d = a<sub>2<\/sub> \u2013 a<sub>1<\/sub> = \\(\\frac{1-3 b}{3}-\\frac{1}{3}\\)<br \/>\n= \\(\\frac{1-3 b-1}{3}=\\frac{-3 b}{3}\\) = -b<\/p>\n<p><strong>Question 6. What is the common difference of an A.P. in which a<sub>21<\/sub> \u2013 a<sub>7<\/sub> = 84? (2017OD )<\/strong><br \/>\nSolution:<br \/>\na<sub>21<\/sub> \u2013 a<sub>7<\/sub> = 84 \u2026[Given<br \/>\n\u2234 (a + 20d) \u2013 (a + 6d) = 84 \u2026[a<sub>n<\/sub> = a + (n \u2013 1)d<br \/>\n20d \u2013 6d = 84<br \/>\n14d = 84 \u21d2 d \\(\\frac{84}{14}\\) = 6<\/p>\n<p><strong>Question 7. Find the 9th term from the end (towards the first term) of the A.P.    5,9,13, \u2026, 185. (2016D)<\/strong><br \/>\nSolution: Here First term, a = 5<br \/>\nCommon difference, d = 9 \u2013 5 = 4<br \/>\nLast term, 1 = 185<br \/>\nn<sup>th<\/sup> term from the end = l \u2013 (n \u2013 1)d<br \/>\n9<sup>th<\/sup> term from the end = 185 \u2013 (9 \u2013 1)4<br \/>\n= 185 \u2013 8 \u00d7 4 = 185 \u2013 32 = 153<\/p>\n<p style=\"text-align: center;\"><strong>Also Check: Important Questions for Class 10 Maths Chapter 1 \u2013 Real Numbers<\/strong><\/p>\n<h3><span class=\"ez-toc-section\" id=\"Arithmetic_Progressions_Class_10_Important_Questions_Short_Answer-I_2_Marks\"><\/span>Arithmetic Progressions Class 10 Important Questions Short Answer-I (2 Marks)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 8. The angles of a triangle are in A.P., the least being half the greatest. Find the angles. (2011D)<\/strong><br \/>\nSolution:<br \/>\nLet the angles be a \u2013 d, a, a + d; a &gt; 0, d &gt; 0<br \/>\n\u2235 Sum of angles = 180\u00b0<br \/>\n\u2234 a \u2013 d + a + a + d = 180\u00b0<br \/>\n\u21d2 3a = 180\u00b0 \u2234 a = 60\u00b0 \u2026(i)<br \/>\nBy the given condition<br \/>\na \u2013 d = \\(\\frac{a+d}{2}\\)<br \/>\n\u21d2 2 = 2a \u2013 2d = a + d<br \/>\n\u21d2 2a \u2013 a = d + 2d \u21d2 a = 3d<br \/>\n\u21d2 d = \\(\\frac{a}{3}=\\frac{60^{\\circ}}{3}\\) = 20\u00b0 \u2026 [From (i)<br \/>\n\u2234 Angles are: 60\u00b0 \u2013 20\u00b0, 60\u00b0, 60\u00b0 + 20\u00b0<br \/>\ni.e., 40\u00b0, 60\u00b0, 80\u00b0<\/p>\n<p><strong>Question 9. Find whether -150 is a term of the A.P. 17, 12, 7, 2, \u2026 ? (2011D)<\/strong><br \/>\nSolution:<br \/>\nGiven: 1<sup>st<\/sup> term, a = 17<br \/>\nCommon difference, d = 12 \u2013 17 = -5<br \/>\nn<sup>th<\/sup> term, a<sub>n<\/sub> = \u2013 150 (Let)<br \/>\n\u2234 a + (n \u2013 1) d = \u2013 150<br \/>\n17 + (n \u2013 1)(-5) = \u2013 150<br \/>\n(n \u2013 1) (-5) = \u2013 150 \u2013 17 = \u2013 167<br \/>\n(n \u2212 1) = \\(\\frac{-167}{-5}\\)<br \/>\nn = \\(\\frac{167}{5}\\) + 1 = \\(\\frac{167+5}{5}=\\frac{172}{5}\\)<br \/>\nn = \\(\\frac{172}{5}\\) \u2026[Being not a natural number<br \/>\n\u2234 -150 is not a term of given A.P.<\/p>\n<p><strong>Question 10. Which term of the progression 4, 9, 14, 19, \u2026 is 109? (2011D)<\/strong><br \/>\nSolution:<br \/>\nHere, d = 9 -4 = 14 -9 = 19 \u2013 14 = 5<br \/>\n\u2234 Difference between consecutive terms is constant.<br \/>\nHence it is an A.P.<br \/>\nGiven: First term, a = 4, d = 5, a<sub>n<\/sub> = 109 (Let)<br \/>\n\u2234 a<sub>n<\/sub> = a + (n \u2013 1) d \u2026 [General term of A.P.<br \/>\n\u2234 109 = 4 + (n \u2013 1) 5<br \/>\n\u21d2 109 \u2013 4 = (n \u2013 1) 5<br \/>\n\u21d2 105 = 5(n \u2212 1) \u21d2 n \u2013 1 = \\(\\frac{105}{5}\\) = 21<br \/>\n\u21d2 n = 21 + 1 = 22 \u2234 109 is the 22<sup>nd<\/sup> term<\/p>\n<p><strong>Question 11. Which term of the progression 20, 192, 183, 17 \u2026 is the first negative term? (2017OD)<\/strong><br \/>\nSolution:<br \/>\nGiven: A.P.: 20, \\(\\frac{77}{4}, \\frac{37}{4}, \\frac{71}{4}\\)<br \/>\nHere a = 20, d = \\(\\frac{77-80}{4}=-\\frac{3}{4}\\)<br \/>\nFor first negative term, a<sub>n<\/sub> &lt; 0<br \/>\n\u21d2 a + (n \u2212 1)d &lt; 0 \u21d2 20 + (n \u2212 1)(-\\(\\frac{3}{4}\\)) &lt; 0<br \/>\n\u21d2 \u2013\\(\\frac{3}{4}\\)(n \u2212 1) &lt; -20 \u21d2 3(1 \u2013 1) &gt; 80<br \/>\n\u21d2 3n \u2013 3 &gt; 80 \u21d2 3n &gt; 83<br \/>\nn &gt; \\(\\frac{83}{4}\\) \u21d2 n &gt; 27.5<br \/>\n\u2234 Its negative term is 28th term.<\/p>\n<p><strong>Question 12. The 4<sup>th<\/sup> term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11<sup>th<\/sup>term. (2016OD)<\/strong><br \/>\nSolution:<br \/>\nLet 1<sup>st<\/sup> term = a, Common difference = d<br \/>\na<sub>4<\/sub> = 0 a + 3d = 0 \u21d2 a = -3d \u2026 (i)<br \/>\nTo prove: a<sub>25<\/sub> = 3 \u00d7 a<sub>11<\/sub><br \/>\na + 24d = 3(a + 10d) \u2026[From (i)<br \/>\n\u21d2 -3d + 24d = 3(-3d + 10d)<br \/>\n\u21d2 21d = 21d<br \/>\nFrom above, a<sub>25<\/sub> = 3(a<sub>11<\/sub>) (Hence proved)<\/p>\n<p><strong>Question 13. The 7<sup>th<\/sup> term of an A.P. is 20 and its 13<sup>th<\/sup> term is 32. Find the A.P. (2012OD)<\/strong><br \/>\nSolution:<br \/>\nLet a be the 1<sup>th<\/sup> term and d be the common difference.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130856\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-2.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 2\" width=\"394\" height=\"241\" \/><\/p>\n<p><strong>Question 14. Find 10th term from end of the A.P. 4,9, 14, \u2026, 254. (2011OD)<\/strong><br \/>\nSolution:<br \/>\nCommon difference d = 9 \u2013 4<br \/>\n= 14 \u2013 9 = 5<br \/>\nGiven: Last term, l = 254, n = 10<br \/>\nn<sup>th<\/sup> term from the end = l \u2013 (n \u2013 1) d<br \/>\n\u2234 10<sup>th<\/sup> term from the end = 254 \u2013 (10 \u2013 1) \u00d7 5<br \/>\n= 254 \u2013 45 = 209<\/p>\n<p><strong>Question 15. Find how many two-digit numbers are divisible by 6? (2011OD)<\/strong><br \/>\nSolution:<br \/>\n12, 18, 24, \u2026,96<br \/>\nHere, a = 12, d = 18 \u2013 12 = 6, a<sub>n<\/sub> = 96<br \/>\na + (n \u2013 1)d = a<sub>n<\/sub><br \/>\n\u2234 12 + (n \u2013 1)6 = 96<br \/>\n\u21d2 (n \u2212 1)6 = 96 \u2013 12 = 84<br \/>\n\u21d2 n \u2013 1 = \\(\\frac{84}{6}\\) = 14<br \/>\n\u21d2 n = 14 + 1 = 15<br \/>\n\u2234 There are 15 two-digit numbers divisible by 6.<\/p>\n<p><strong>Question 16. How many natural numbers are there between 200 and 500, which are divisible by 7? (2011OD)<\/strong><br \/>\nSolution:<br \/>\n203, 210, 217, \u2026, 497<br \/>\nHere a = 203, d = 210 \u2013 203 = 7, a<sub>n<\/sub> = 497<br \/>\n\u2234 a + (n \u2013 1) d = a<sub>n<\/sub><br \/>\n203 + (n \u2013 1) 7 = 497<br \/>\n(n \u2013 1) 7 = 497 \u2013 203 = 294<br \/>\nn \u2013 1 = \\(\\frac{294}{7}\\) = 42 \u2234 n = 42 + 1 = 43<br \/>\n\u2234 There are 43 natural nos. between 200 and 500 which are divisible by 7.<\/p>\n<p><strong>Question 17. How many two-digit numbers are divisible by 3? (2012OD)<\/strong><br \/>\nSolution:<br \/>\nTwo-digit numbers divisible by 3 are:<br \/>\n12, 15, 18, \u2026, 99<br \/>\nHere, a = 12, d = 15 \u2013 12 = 3, a<sub>n<\/sub> = 99<br \/>\n\u2234 a + (n \u2013 1)d = a<sub>n<\/sub><br \/>\n12 + (n \u2013 1) (3) = 99<br \/>\n(n \u2013 1) (3) = 99 \u2013 12 = 87<br \/>\nn \u2013 1 = \\(\\frac{87}{3}\\) = 29<br \/>\n\u2234 n = 29 + 1 = 30<br \/>\n\u2234 There are 30 such numbers.<\/p>\n<p><strong>Question 18. How many three-digit natural numbers are divisible by 7? (2013D)<\/strong><br \/>\nSolution:<br \/>\n\u201c3 digits nos.\u201d are 100, 101, 102, \u2026, 999<br \/>\n3 digits nos. \u201cdivisible by 7\u201d are:<br \/>\n105, 112, 119, 126, \u2026, 994<br \/>\na = 105, d = 7, a<sub>n<\/sub> = 994, n = ?<br \/>\nAs a + (n \u2013 1)d = 994 = a<sub>n<\/sub><br \/>\n\u2234 105 + (n \u2013 1)7 = 994<br \/>\n(n \u2212 1)7 = 994 \u2013 105<br \/>\n(n \u2013 1) = \\(\\frac{889}{7}\\) = 127<br \/>\n\u2234 n = 127 + 1 = 128<\/p>\n<p><strong>Question 19. Find the number of all three-digit natural numbers which are divisible by 9. (2013OD)<\/strong><br \/>\nSolution:<br \/>\nTo find: Number of terms of A.P., i.e., n.<br \/>\nA.P. = 108 + 117 + 126 + \u2026 + 999<br \/>\n1<sup>st<\/sup> term, a = 108<br \/>\nCommon difference, d = 117 \u2013 108 = 9<br \/>\na<sub>n<\/sub> = 999<br \/>\na + (n \u2013 1)d = a<sub>n<\/sub><br \/>\n\u2234 108 + (n \u2013 1) 9 = 999<br \/>\n\u21d2 (n \u2212 1) 9 = 999 \u2013 108 = 891<br \/>\n\u21d2 (n \u2212 1) = \\(\\frac{891}{9}\\) = 99<br \/>\n\u2234 n = 99 + 1 = 100<\/p>\n<p><strong>Question 20. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. (2014OD)<\/strong><br \/>\nSolution:<br \/>\nNumbers divisible by both 2 and 5 are 110, 120, 130, \u2026, 990.<br \/>\nHere a = 110, d = 120 \u2013 110 = 10, a<sub>n<\/sub> = 990<br \/>\nAs a + (n \u2013 1)d = a<sub>n<\/sub> = 990<br \/>\n110 + (n \u2013 1)(10) = 990<br \/>\n(n \u2013 1)(10) = 990 \u2013 110 = 880<br \/>\n(n \u2212 1) = \\(\\frac{294}{7}\\) = 88<br \/>\n\u2234 n = 88 + 1 = 89<\/p>\n<p><strong>Question 21. Find the middle term of the A.P. 6, 13, 20, \u2026, 216. (2015D)<\/strong><br \/>\nSolution:<br \/>\nThe given A.P. is 6, 13, 20, \u2026, 216<br \/>\nLet n be the number of terms, d = 7, a = 6, a<sub>n<\/sub> = 216<br \/>\na<sub>n<\/sub> = a + (n \u2013 1)d<br \/>\n\u2234 216 = 6 + (n \u2013 1).7<br \/>\n216 \u2013 6 = (n \u2013 1)7<br \/>\n\\(\\frac{210}{7}\\) = n \u2013 1 \u21d2 30 + 1 = n<br \/>\n\u21d2 n = 31<br \/>\nMiddle term = \\(\\left(\\frac{n+1}{2}\\right)^{t h}\\) term<br \/>\n= \\(\\left(\\frac{31+1}{2}\\right)=\\left(\\frac{32}{2}\\right)\\) = 16<sup>th<\/sup> term of the A.P.<br \/>\n\u2234 a<sub>16<\/sub> = a + 15d = 6 + 15 \u00d7 7 = 111<\/p>\n<p><strong>Question 22. Find the middle term of the A.P. 213, 205, 197, \u2026 37. (2015D)<\/strong><br \/>\nSolution:<br \/>\nA.P. : 213, 205, 197, \u2026\u2026\u2026\u2026. 37.<br \/>\nLet a and d be the first term and common difference of A.P. respectively,<br \/>\nHere a = 213, d = -8, a<sub>n<\/sub> = 37, where n is the number of terms.<br \/>\na<sub>n<\/sub> = a + (n \u2013 1)d<br \/>\n\u2234 37 = 213 + (n \u2013 1) (-8)<br \/>\n\\(\\frac{-176}{-8}\\) = n \u2013 1 \u21d2 n = 23<br \/>\n\u2234 Middle term = \\(\\left(\\frac{n+1}{2}\\right)^{\\text { th }}=\\left(\\frac{23+1}{2}\\right)\\)<br \/>\n= 12<sup>th<\/sup> term<br \/>\n\u2234 a<sub>12<\/sub> = a + 11(d) = 213 + 11(-8) = 125<\/p>\n<p><strong>Question 23. How many terms of the A.P. 27, 24, 21, \u2026 should be taken so that their sum is zero? (2016D)<\/strong><br \/>\nSolution:<br \/>\nHere, 1<sup>st<\/sup> term, a = 27<br \/>\nCommon difference, d = 24 \u2013 27 = -3<br \/>\nGiven: S<sub>n<\/sub> = 0<br \/>\n\u21d2 \\(\\frac{n}{2}\\)[2a + (n \u2212 1)d] = 0<br \/>\n\u21d2 \\(\\frac{n}{2}\\)[2(27) + (n \u2212 1)(-3)] = 0<br \/>\n\u21d2 n(54 \u2013 3n + 3) = 0<br \/>\n\u21d2 n(57 \u2013 3n) = 0<br \/>\n\u21d2 n = 0 or (57 \u2013 3n) = 0<br \/>\n\u21d2 -3n = -57<br \/>\n\u21d2 n = 19<br \/>\nSince n, i.e., number of terms cannot be zero.<br \/>\n\u2234 Number of terms = 19<\/p>\n<p><strong>Question 24. How many terms of the A.P. 65, 60, 55, \u2026 be taken so that their sum is zero? (2016D)<\/strong><br \/>\nSolution:<br \/>\n1<sup>st<\/sup> term, a = 65<br \/>\nCommon difference, d = 60 \u2013 65 = -5<br \/>\nS<sub>n<\/sub> = 0 \u2026[Given<br \/>\n\u21d2 \\(\\frac{n}{2}\\)[2a + (n \u2212 1)d] = 0<br \/>\n\u21d2 \\(\\frac{n}{2}\\)[2(65) + (n \u2212 1)(-5)] = 0<br \/>\n\u21d2 n(130 \u2013 5n + 5) = 0<br \/>\n\u21d2 n(135 \u2013 5n) = 0<br \/>\n\u21d2 n = 0 or 135 \u2013 5n = 0<br \/>\n-5n = -135<br \/>\n\u21d2 n = 27<br \/>\nSince n, i.e., number of terms can not be zero.<br \/>\n\u2234 Number of terms = 27<\/p>\n<p><strong>Question 25. Find the sum of the first 25 terms of an A.P. whose n<sup>th<\/sup> term is given by t<sub>n<\/sub> = 2 \u2013 3n. (2012D)<\/strong><br \/>\nSolution:<br \/>\nGiven: t<sub>n<\/sub> = 2 \u2013 3n<br \/>\nWhen n = 1, t<sub>1<\/sub> = 2 \u2013 3(1) = -1 ..(i)<br \/>\nWhen n = 25, t<sub>25<\/sub> = 2 \u2013 3(25) = -73 \u2026(ii)<br \/>\nAs S<sub>n<\/sub> = \\(\\frac{n}{2}\\)[t<sub>1<\/sub> + t<sub>n<\/sub>], n = 25<br \/>\n\u2234 S<sub>25<\/sub> = \\(\\frac{25}{2}\\)[-1 + (-73)] .. [From (1) and (ii)<br \/>\n= \\(\\frac{25 \\times(-74)}{2}\\) = 25 \u00d7 (-37) = -925<\/p>\n<p><strong>Question 26. The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference. (2014D)<\/strong><br \/>\nSolution:<br \/>\nHere, a<sub>n<\/sub> = 45, S<sub>n<\/sub> = 400, a = 5, n = ?, d = ?<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130857\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-3.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 3\" width=\"392\" height=\"218\" \/><\/p>\n<p><strong>Question 27. The first and the last terms of an AP are 8 and 65 respectively. If the sum of all its terms is 730, find its common difference. (2014D)<\/strong><br \/>\nSolution:<br \/>\nHere a, = a = 8; a<sub>n<\/sub> = 65<br \/>\nGiven: S<sub>n<\/sub> = 730<br \/>\n\u21d2 \\(\\frac{n}{2}\\)(8 + 65) = 730 \u2026..[S<sub>n<\/sub> = \\(\\frac{n}{2}\\)(a<sub>1<\/sub> + a<sub>n<\/sub>)<br \/>\n\\(\\frac{n}{2}\\)(73) = 730<br \/>\nn = 730 \u00d7 \\(\\frac{2}{73}\\) = 20<br \/>\nNow, a<sub>n<\/sub> = a + (n \u2212 1)d = 65<br \/>\n8 +(20 \u2013 1)d = 65 \u21d2 19d = 65 \u2013 8 = 57<br \/>\n\u2234 d = 3<\/p>\n<p><strong>Question 28. In an AP, if S<sub>5<\/sub> + S<sub>7<\/sub> = 167 and S<sub>10<\/sub> = 235, then find the AP, where s, denotes the sum of its first n terms. (2015OD)<\/strong><br \/>\nSolution:<br \/>\nGiven: S<sub>5<\/sub> + S<sub>7<\/sub> = 167<br \/>\n\u21d2 \\(\\frac{5}{2}\\)[2a + (5 \u2013 1)d] + \\(\\frac{7}{2}\\) [2a + (7 \u2013 1)d] = 167 \u2026 [S&lt;sub&lt;n = \\(\\frac{n}{2}\\) (2a + (n \u2013 1)d)<br \/>\n\u21d2 \\(\\frac{5}{2}\\)[2a + 4d] + \\(\\frac{7}{2}\\)[2a + 6d] = 167<br \/>\n\u21d2 5(a + 2d) + 7(a + 3d) = 167<br \/>\n\u21d2 5a + 10d + 7a + 210 = 167<br \/>\n\u21d2 12a + 31d = 167<br \/>\nNow, S<sub>10<\/sub> = \\(\\frac{10}{2}\\) (2a + (10 \u2013 1)d) = 235<br \/>\n\u21d2 5[2a + 9d] = 235<br \/>\n\u21d2 10a + 45d = 235<br \/>\nSolving (i) and (ii), we get a = 1 and d = 5<br \/>\na<sub>1<\/sub> = 1<br \/>\na<sub>2<\/sub> = a + d \u21d2 1 + 5 = 6<br \/>\na<sub>3<\/sub> = a + 2d \u21d2 1 + 10 = 11<br \/>\nHence A.P. is 1, 6, 11\u2026<\/p>\n<p><strong>Question 29. Find the sum of all three digit natural numbers, which are multiples of 11. (2012D)<\/strong><br \/>\nSolution:<br \/>\nTo find: 110 + 121 + 132 + \u2026 + 990<br \/>\nHere a = 110, d = 121- 110 = 11, a<sub>n<\/sub> = 990<br \/>\n\u2234 a + (n \u2013 1)d = 990<br \/>\n110 + (n \u2013 1).11 = 990<br \/>\n(n \u2013 1). 11 = 990 \u2013 110 = 880<br \/>\n(n \u2013 1) = 880 = 80<br \/>\nn = 80 + 1 = 81<br \/>\nAs S<sub>n<\/sub> = \\(\\frac{n}{2}\\) (a<sub>1<\/sub> + a<sub>n<\/sub>)<br \/>\n\u2234 S<sub>81<\/sub> = \\(\\frac{81}{2}\\) (110 + 990)<br \/>\n= \\(\\frac{81}{2}\\) (1100) = 81 \u00d7 550 = 44,550<\/p>\n<h3><span class=\"ez-toc-section\" id=\"Arithmetic_Progressions_Class_10_Important_Questions_Short_Answer_-_II_3_Marks\"><\/span>Arithmetic Progressions Class 10 Important Questions Short Answer \u2013 II (3 Marks)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 30. Which term of the A.P. 3, 14, 25, 36, \u2026 will be 99 more than its 25th term? (2011OD)<\/strong><br \/>\nSolution:<br \/>\nLet the required term be n<sup>th<\/sup> term, i.e., a<sub>n<\/sub><br \/>\nHere, d = 14 \u2013 3 = 11, a = 3<br \/>\nAccording to the Question, a<sub>n<\/sub> = 99 + a<sub>25<\/sub><br \/>\n\u2234 a + (n \u2013 1) d = 99 + a + 24d<br \/>\n\u21d2 (n \u2013 1) (11) = 99 + 24 (11)<br \/>\n(n \u2013 1) (11) = 11 (9 + 24)<br \/>\nn \u2013 1 = 33<br \/>\nn = 33 + 1 = 34<br \/>\n\u2234 34<sup>th<\/sup> term is 99 more than its 25th term.<\/p>\n<p><strong>Question 31. Determine the A.P. whose fourth term is 18 and the difference of the ninth term from the fifteenth term is 30. (2011D)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130858\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-4.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 4\" width=\"396\" height=\"131\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130859\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-5.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 5\" width=\"352\" height=\"97\" \/><\/p>\n<p><strong>Question 32. The 19th term of an AP is equal to three times its 6<sup>th<\/sup> term. If its 9<sup>th<\/sup> term is 19, find the A.P. (2013OD)<\/strong><br \/>\nSolution:<br \/>\nGiven: a<sub>19<\/sub> = 3(a<sub>6<\/sub>)<br \/>\n\u21d2 a + 18d = 3(a + 5d)<br \/>\na + 18d = 3a + 15d<br \/>\n18d \u2013 15d = 3a \u2013 a<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130860\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-6.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 6\" width=\"392\" height=\"266\" \/><\/p>\n<p><strong>Question 33. The 9th term of an A.P. is equal to 6 times its second term. If its 5<sup>th<\/sup> term is 22, find the A.P. (2013OD)<\/strong><br \/>\nSolution:<br \/>\n9<sup>th<\/sup> term = 6 (2<sup>nd<\/sup> term)<br \/>\n\u2234 a +8d = 6 (a + d) \u2026[As a<sub>n<\/sub> = a + (n \u2013 1)d<br \/>\na + 8d = 6a + 6d<br \/>\n8d \u2013 6d = 6a \u2013 a<br \/>\n2d = 5a<br \/>\n\u21d2 d = \\(\\frac{5 a}{2}\\) \u2026(i)<br \/>\nNow, a<sub>5<\/sub> = 22<br \/>\na + 4d = 22<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130861\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-7.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 7\" width=\"394\" height=\"190\" \/><\/p>\n<p><strong>Question 34. The sum of the 5th and the 9th terms of an AP is 30. If its 25<sup>th<\/sup> term is three times its 8<sup>th<\/sup> term, find the AP. (2014OD)<\/strong><br \/>\nSolution:<br \/>\na<sub>5<\/sub> + a<sub>9<\/sub> = 30 \u2026 [Given<br \/>\na + 4d + a + 8d = 30 \u2026[\u2235 a<sub>n<\/sub> = a + (n \u2013 1)d<br \/>\n2a + 12d = 30 \u21d2 a + 6d = 15 \u2026[Dividing by 2<br \/>\na = 15 \u2013 6d \u2026(i)<br \/>\nNow, a<sub>52<\/sub> = 3(a<sub>8<\/sub>)<br \/>\na + 24d = 3(a + 7d)<br \/>\n15 \u2013 6d + 240 = 3(15 \u2013 6d + 7d) \u2026[From (i)<br \/>\n15 + 18d = 3(15 + d)<br \/>\n15 + 18d = 45 + 3d<br \/>\n18d \u2013 3d = 45 \u2013 15<br \/>\n15d = 30 \u2234 d = \\(\\frac{30}{15}\\) = 2<br \/>\nFrom (i), a = 15 \u2013 6(2) = 15 \u2013 12 = 3<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130862\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-8.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 8\" width=\"371\" height=\"52\" \/><\/p>\n<p><strong>Question 35. If the seventh term of an AP is \\(\\frac{1}{9}\\) and its ninth term is \\(\\frac{1}{7}\\), find its 63<sup>rd<\/sup> term. (2014D)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130863\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-9.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 9\" width=\"395\" height=\"404\" \/><\/p>\n<p><strong>Question 36. Find the value of the middle term of the following A.P.: -6, -2, 2, \u2026, 58. (2011D)<\/strong><br \/>\nSolution:<br \/>\nHere a = -6, d = -2 -(-6) = 4, a<sub>n<\/sub> = 58<br \/>\nAs we know, a + (n \u2013 1) d = 58<br \/>\n\u2234 -6 + (n \u2013 1) 4 = 58<br \/>\n\u21d2 (n \u2013 1) 4 = 58 + 6 = 64<br \/>\n\u21d2 (n \u2013 1) = \\(\\frac{64}{4}\\) = 16<br \/>\n\u21d2 n = 16 + 1 = 17 (odd)<br \/>\nMiddle term = \\(\\left(\\frac{n+1}{2}\\right)^{t h}\\) term<br \/>\n\\(\\left(\\frac{17+1}{2}\\right)^{t h}\\) term = 9<sup>th<\/sup> term<br \/>\n\u2234 a<sub>9<\/sub> = a + 8d = -6 + 8 (4) = -6 + 32 = 26<br \/>\n\u2234 Middle term = 26<\/p>\n<p><strong>Question 37. Find the number of terms of the AP 18, 151\/2, 13,\u2026, -491\/2 and find the sum of all its terms.<\/strong><br \/>\nSolution:<br \/>\nHere 1<sup>st<\/sup> term, a = 18<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130864\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-10.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 10\" width=\"278\" height=\"125\" \/><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130865\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-11.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 11\" width=\"396\" height=\"343\" \/><\/p>\n<p><strong>Question 38. The 14<sup>th<\/sup> term of an AP is twice its g<sup>th<\/sup> term. If its 6<sup>th<\/sup> term is -8, then find the sum of its first 20 terms. (2015OD)<\/strong><br \/>\nSolution:<br \/>\nLet a = First term, d = Common difference<br \/>\na<sub>14<\/sub> = 2.a<sub>8<\/sub> \u2026 [Given<br \/>\n\u21d2 a + 13d = 2 (a + 7d) ..[\u2235 a, = a + (n \u2013 1)d<br \/>\n\u21d2 a + 13d = 2a + 14d<br \/>\n\u21d2 1a \u2013 2a = 14d \u2013 13d<br \/>\n\u21d2 -1a = d \u21d2 a = -d \u2026 (i)<br \/>\na<sub>6<\/sub> = -8 \u2026[Given<br \/>\n\u21d2 -8 = a + 5d<br \/>\n\u21d2 -d + 5d = -8 \u2026[From (i)<br \/>\n\u21d2 4d = -8 \u21d2 d = -2<br \/>\nValue of d put in equation (i), we get<br \/>\na = -d \u21d2 a=-(-2)<br \/>\nNow, a = 2, d = -2<br \/>\nNow, Sum of first 20 terms,<br \/>\nS<sub>20<\/sub> = \\(\\frac{20}{2}\\)[2 \u00d7 2 + (20 \u2013 1)(-2)] \u2026[S<sub>n<\/sub> = \\(\\frac{n}{2}\\) (2a + (n \u2013 1)d)<br \/>\nS<sub>20<\/sub> = 10[4 + 19(-2)]\nS<sub>20<\/sub> = 10[4 \u2013 38] = -340<\/p>\n<p><strong>Question 39. The 13th term of an AP is four times its 3rd term. If its fifth term is 16, then find the sum of its first ten terms. (2015OD)<\/strong><br \/>\nSolution:<br \/>\na<sub>13<\/sub> = 4a<sub>3<\/sub> \u2026 [Given<br \/>\n\u21d2 a + 12d = 4(a + 2d) \u2026[\u2235 a<sub>n<\/sub> = a + (n \u2013 1)d<br \/>\n\u21d2 a + 12d \u2013 4a \u2013 8d = 0<br \/>\n\u21d2 4d = 3a \u21d2 d = \\(\\frac{3 a}{4}\\) \u2026(i)<br \/>\na<sub>5<\/sub> = 16 \u2026 [Given<br \/>\n\u21d2 a + 4d = 16<br \/>\n\u21d2 a +4(\\(\\frac{3 a}{4}\\)) = 16 \u2026 [From (i)<br \/>\n\u21d2 a + 3a = 16<br \/>\n\u21d2 4a = 16 \u21d2 a = \\(\\frac{16}{4}\\) = 4 ..(ii)<br \/>\nPutting a = 4 in (i), we get a = 3<br \/>\n\u21d2 d = \\(\\frac{3 \\times 4}{4}\\) = 3 \u2026 [From (ii)<br \/>\n\u2234 S<sub>n<\/sub> = \\(\\frac{n}{2}\\) [2a + (n \u2013 1)d]\nS<sub>10<\/sub> = \\(\\frac{10}{2}\\) [2(4) + (10 \u2013 1)(3)] \u2026[n = 10 (Given)<br \/>\nS<sub>10<\/sub> = 5 (8 + 27) \u21d2 5(35) = 175<\/p>\n<p><strong>Question 40. If the sum of first 7 terms of an A.P is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P. (2016D)<\/strong><br \/>\nSolution:<br \/>\nLet 1<sup>st<\/sup> term = a, Common difference = d<br \/>\nGiven: S<sub>7<\/sub> = 49, S<sub>17<\/sub> = 289<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130866\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-12.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 12\" width=\"400\" height=\"495\" \/><\/p>\n<p><strong>Question 41. The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 400. Find the number of terms and the common difference of the A.P. (2017OD)<\/strong><br \/>\nSolution:<br \/>\nFirst term, a = 5, Last term, a<sub>n<\/sub> = 45<br \/>\nLet the number of terms = n<br \/>\nS<sub>n<\/sub> = 400<br \/>\n\\(\\frac{n}{2}\\) (a + a<sub>n<\/sub>) = 400<br \/>\n\\(\\frac{n}{2}\\)(5 + 45) = 400<br \/>\n\\(\\frac{n}{2}\\) (50) = 400<br \/>\nn = \\(\\frac{400}{25}\\) = 16 = Number of terms<br \/>\nNow, a<sub>n<\/sub> = 45<br \/>\na + (n \u2013 1)d = a<sub>n<\/sub><br \/>\n5+ (16 \u2013 1)d = 45<br \/>\n15d = 45 \u2013 5 \u2234 d = \\(\\frac{40}{15}=\\frac{8}{3}\\)<\/p>\n<p><strong>Question 42. The n<sup>th<\/sup> term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P. (2013D)<\/strong><br \/>\nSolution:<br \/>\nWe have, a<sub>n<\/sub> = -4n + 15<br \/>\nPut n = 1, a<sub>1<\/sub> = -4(1) + 15 = 11<br \/>\nPut n=2, a<sub>2<\/sub> = -4(2) + 15 = 7<br \/>\n\u2234 d = a<sub>2<\/sub> \u2013 a<sub>1<\/sub> = 7 \u2013 11 = -4<br \/>\nAs S<sub>n<\/sub> = \\(\\frac{n}{2}\\) (2a + (n \u2013 1)d]\n\u2234 S<sub>20<\/sub> = \\(\\frac{20}{2}\\) [2(11) + (20 \u2013 1)(-4))\u2026 [n = 20 (Given)<br \/>\n= 10 (22 \u2013 76)<br \/>\n= 10 (-54) = -540<\/p>\n<p><strong>Question 43. The sum of first n terms of an AP is 3n<sup>2<\/sup> + 4n. Find the 25<sup>th<\/sup> term of this AP. (2013D)<\/strong><br \/>\nSolution:<br \/>\nWe have, S<sub>n<\/sub> = 3n<sup>2<\/sup> + 4n<br \/>\nPut n = 25,<br \/>\nS<sub>25<\/sub> = 3(25)<sup>2<\/sup> + 4(25)<br \/>\n= 3(625) + 100<br \/>\n= 1875 + 100 = 1975<br \/>\nPut n = 24,<br \/>\nS<sub>24<\/sub> = 3(24)<sup>2<\/sup> + 4(24)<br \/>\n= 3(576) + 96<br \/>\n= 1728 + 96 = 1824<br \/>\n\u2234 25<sup>th<\/sup> term = S<sub>25<\/sub> \u2013 S<sub>24<\/sub><br \/>\n= 1975 \u2013 1824 = 151<\/p>\n<p><strong>Question 44. The sum of the first seven terms of an AP is 182. If its 4<sup>th<\/sup> and the 17<sup>th<\/sup> terms are in the ratio 1 : 5, find the AP. (2014OD)<\/strong><br \/>\nSolution:<br \/>\nS<sub>7<\/sub> = 182 \u2026[Given<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130867\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-13.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 13\" width=\"398\" height=\"461\" \/><br \/>\n\u2234 a = 2, d = 8<br \/>\n\u2234AP is 2, 10, 18, 26, 34, \u2026<\/p>\n<p><strong>Question 45. If S<sub>n<\/sub>, denotes the sum of first n terms of an A.P., prove that S<sub>12<\/sub> = 3(S<sub>8<\/sub> \u2013 S<sub>4<\/sub>). (2015D)<\/strong><br \/>\nSolution:<br \/>\nLet a be the first term and d be the common difference of A.P.<br \/>\nS<sub>n<\/sub> = \\(\\frac{n}{2}\\) (2a + (n \u2212 1)d)<br \/>\n\u2234 S<sub>12<\/sub> = \\(\\frac{12}{2}\\) (2a + (12 \u2013 1)d)<br \/>\nS<sub>12<\/sub> = 6 [2a + 11d] = 124 + 66d \u2026(i)<br \/>\n\u2234 S<sub>8<\/sub> = \\(\\frac{8n}{2}\\) (2a + (8 \u2013 1)d)<br \/>\nS<sub>8<\/sub> = 4[2a + 7d] = 8a + 28d \u2026 (ii)<br \/>\n\u2234 S<sub>4<\/sub> = \\(\\frac{4}{2}\\) (2a + (4 \u2013 1)d)<br \/>\nS<sub>4<\/sub> = 2[2a + 3d) = 4a + 6d \u2026(iii)<br \/>\nNow, S<sub>12<\/sub> = 3(S<sub>8<\/sub> \u2013 S<sub>4<\/sub>)<br \/>\n12a + 660 = 3(8a + 28d \u2013 4a \u2013 6d) \u2026 [From (i), (ii) &amp; (iii)<br \/>\n12a + 660 = 3(4a + 22d)<br \/>\n12a + 660 = 12a + 66d \u2026Hence proved<\/p>\n<p><strong>Question 46. If the sum of the first n terms of an A.P. is \\(\\frac{1}{2}\\) (3n<sup>2<\/sup> + 7n), then find its n<sup>th<\/sup> term. Hence write its 20<sup>th<\/sup> (2015D)<\/strong><br \/>\nSolution:<br \/>\nS<sub>n<\/sub> = \\(\\frac{1}{2}\\) (3n<sup>2<\/sup> + 7n) \u2026[Given<br \/>\nPut n = 1<br \/>\nS<sub>1<\/sub> = \\(\\frac{1}{2}\\)[3(1)<sub>2<\/sub> + 7(1)] = \\(\\frac{1}{2}\\) [3 + 7) = \\(\\frac{1}{2}\\) (10) = 5<br \/>\nPut n = 2<br \/>\nS<sub>2<\/sub> = \\(\\frac{1}{2}\\) [3(2)<sup>2<\/sup> + 7(2)] = \\(\\frac{1}{2}\\)[3(4) + 7(2)]\n= S<sub>2<\/sub> = \\(\\frac{1}{2}\\)(12 + 14) = \\(\\frac{1}{2}\\) (26) = 13<br \/>\nNow we know,<br \/>\nS<sub>1<\/sub> = 27 and a<sub>2<\/sub> = S<sub>2<\/sub> \u2013 S<sub>1<\/sub><br \/>\n\u2234 a<sub>1<\/sub> = 5 = 13 \u2013 5 = 8<br \/>\nNow we have,<br \/>\na<sub>1<\/sub> = 5, a<sub>2<\/sub> = 8, d = a<sub>2<\/sub> \u2013 a<sub>1<\/sub> = 8 \u2013 5 = 3<br \/>\na<sub>n<\/sub> = a + (n \u2013 1)d = 5 + (n \u2013 1)3<br \/>\n= 5 + 3n \u2013 3 = 3n + 2 (n<sup>th<\/sup> term)<br \/>\n\u2234 20<sup>th<\/sup> term, a<sub>20<\/sub> = (3 \u00d7 20) + 2 = 62<\/p>\n<p><strong>Question 47. If S<sub>n<\/sub> denotes the sum of first n terms of an A.P., prove that S<sub>30<\/sub> = 3[S<sub>20<\/sub> \u2013 S<sub>10<\/sub>]. (2015D)<\/strong><br \/>\nSolution:<br \/>\nLet a be the first term and d be the common difference of the A.P.<br \/>\nS<sub>n<\/sub> = \\(\\frac{n}{2}\\)[2a + (n \u2212 1)d]\nS<sub>30<\/sub> = \\(\\frac{30}{2}\\) [2a + (30 \u2013 1)d] = 15[2a + 29d]\n= 30a + 435d \u2026(i)<br \/>\nS<sub>20<\/sub> = \\(\\frac{20}{2}\\) [2a + (20 \u2013 1)d] = 10[2a + 19d]\n= 20a + 190d \u2026(ii)<br \/>\nS<sub>10<\/sub> = \\(\\frac{10}{2}\\)[2a + (10 \u2013 1)d] = 5[2a + 9d]\n= 10a + 45d \u2026(iii)<br \/>\nTo prove, 3(S<sub>20<\/sub> \u2013 S<sub>10<\/sub>) = S<sub>30<\/sub><br \/>\n= 3(20a + 190d \u2013 10a \u2013 45d) \u2026[From (i), (ii) &amp; (iii)<br \/>\n= 3(10a + 145d)<br \/>\n= 30a + 435d = S<sub>30<\/sub> \u2026Hence Proved<\/p>\n<p><strong>Question 48. If the ratio of the sum of first n terms of two A.Ps is (7n + 1): (4n + 27), find the ratio of their mth terms. (2016OD)<\/strong><br \/>\nSolution:<br \/>\nLet A be first term and D be the common difference of 1<sup>st<\/sup> A.P.<br \/>\nLet a be the first term and d be the common difference of 2<sup>nd<\/sup> A.P.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130868\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-14.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 14\" width=\"399\" height=\"362\" \/><br \/>\n\u2234 Required ratio = (14m \u2013 6) : (8m + 23)<\/p>\n<p><strong>Question 49. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number. (2016OD)<\/strong><br \/>\nSolution:<br \/>\nLet hundred\u2019s place digit = (a \u2013 d)<br \/>\nLet ten\u2019s place digit = a<br \/>\nLet unit\u2019s place digit = a + d<br \/>\nAccording to the Question,<br \/>\na \u2013 d + a + a + d = 15<br \/>\n\u21d2 3a = 15 \u21d2 a = 5<br \/>\nOriginal number<br \/>\n= 100(a \u2013 d) + 10(a) + 1(a + d)<br \/>\n= 100a \u2013 100d + 10a + a + d<br \/>\n= 111a \u2013 99d<br \/>\nReversed number<br \/>\n= 1(a \u2013 d) + 10a + 100(a + d)<br \/>\n= a \u2013 d + 10a + 100a + 100d<br \/>\n= 111a + 99d<br \/>\nNow, Original no. \u2013 Reversed no. = 594<br \/>\n111a \u2013 99d \u2013 (111a + 99d) = 594<br \/>\n-198d = 594 \u21d2 d = \\(\\frac{594}{-198}\\) = -3<br \/>\n\u2234 The Original no. = 111a \u2013 99d<br \/>\n= 111(5) \u2013 99(-3)<br \/>\n= 555 + 297 = 852<\/p>\n<p><strong>Question 50. The sums of first n terms of three arithmetic progressions are S<sub>1<\/sub> S<sub>2<\/sub> and S<sub>3<\/sub> respectively. The first term of each A.P. is 1 and their common differences are 1, 2 and 3 respectively. Prove that S<sub>1<\/sub> + S<sub>3<\/sub> = 2S<sub>2<\/sub>. (2016OD)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130869\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-15.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 15\" width=\"393\" height=\"492\" \/><\/p>\n<p><strong>Question 51. Find the sum of all three digit natural numbers, which are multiples of 9. (2012D)<\/strong><br \/>\nSolution:<br \/>\nTo find: 108 + 117 + 126 + \u2026 + 999<br \/>\n1<sup>st<\/sup> term, a = 108<br \/>\nCommon difference, d = 117 \u2013 108 = 9<br \/>\n\u2234 a + (n \u2013 1)d = a<sub>n<\/sub> = 999<br \/>\n108 + (n \u2013 1). 9 = 999<br \/>\n(n \u2013 1)9 = 999 \u2013 108 = 891<br \/>\n(n \u2013 1) = \\(\\frac{891}{9}\\) = 99<br \/>\nn = 99 + 1 = 100<br \/>\nAs S<sub>n<\/sub>= \\(\\frac{n}{2}\\)(a<sub>1<\/sub> + a<sub>n<\/sub>)<br \/>\n\u2234 S<sub>100<\/sub> = \\(\\[\\frac{100}{2}\\)\\] (108 + 999)<br \/>\n= 50(1107) = 55350<\/p>\n<p><strong>Question 52. Find the sum of all multiples of 7 lying between 500 and 900. (2012OD)<\/strong><br \/>\nSolution:<br \/>\nTo find: 504 + 511 + 518 + \u2026 + 896<br \/>\na = 504, d = 511- 504 = 7, a<sub>n<\/sub> = 896<br \/>\na + (n \u2013 1)d = a<sub>n<\/sub><br \/>\n\u2234 504 + (n \u2013 1)7 = 896<br \/>\n(n \u2013 1)7 = 896 \u2013 504 = 392<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130870\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-16.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 16\" width=\"274\" height=\"220\" \/><\/p>\n<h3><span class=\"ez-toc-section\" id=\"Arithmetic_Progressions_Class_10_Important_Questions_Long_Answer_4_Marks\"><\/span>Arithmetic Progressions Class 10 Important Questions Long Answer (4 Marks)<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 53. If p<sup>th<\/sup>, q<sup>th<\/sup> and r<sup>th<\/sup> terms of an A.P. are a, b, c respectively, then show that (a \u2013 b)r + (b \u2013 c)p+ (c \u2013 a)q = 0. (2011D)<\/strong><br \/>\nSolution:<br \/>\nLet A be the first term and D be the common difference of the given A.P.<br \/>\np<sup>th<\/sup> term = A + (p \u2013 1)D = a \u2026(i)<br \/>\nq<sup>th<\/sup> term = A + (q \u2013 1)D = b \u2026(ii)<br \/>\nr<sup>th<\/sup> term = A + (r \u2013 1)D = c \u2026 (iii)<br \/>\nL.H.S. = (a \u2013 b)r + (b \u2013 c)p + (c \u2013 a)q<br \/>\n= [A + (p \u2013 1)D \u2013 (A + (q \u2013 1)D)]r + [A + (q \u2013 1)D \u2013 (A + (r \u2013 1)D)]p + [A + (r \u2013 1)D \u2013 (A + (p \u2013 1)D)]q<br \/>\n= [(p \u2013 1 \u2013 q + 1)D]r + [(q \u2013 1 \u2013 r + 1)D]p + [(r \u2013 1 \u2013 p + 1)D]q<br \/>\n= D[(p \u2013 q)r + (q \u2013 r)p + (r \u2013 p)q]\n= D[pr \u2013 qr + qp \u2013 rp + rq \u2013 pq]\n= D[0] = 0 = R.H.S.<\/p>\n<p><strong>Question 54. The 17<sup>th<\/sup> term of an AP is 5 more than twice its 8<sup>th<\/sup> term. If the 11<sup>th<\/sup> term of the AP is 43, then find its n<sup>th<\/sup> term. (2012D)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130871\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-17.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 17\" width=\"396\" height=\"145\" \/><br \/>\nFrom (i),<br \/>\na = 2(4) \u2013 5 = 8 \u2013 5 = 3<br \/>\nAs a<sub>n<\/sub> = a + (n \u2013 1) d<br \/>\n\u2234 a<sub>n<\/sub> = 3 + (n \u2013 1) 4 = 3 + 4n \u2013 4<br \/>\na<sub>n<\/sub> = (4n \u2013 1)<\/p>\n<p><strong>Question 55. The 15<sup>th<\/sup> term of an AP is 3 more than twice its 7<sup>th<\/sup> term. If the 10<sup>th<\/sup> term of the AP is 41, then find its n<sup>th<\/sup> term. (2012D)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130872\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-18.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 18\" width=\"393\" height=\"141\" \/><br \/>\nFrom (i),<br \/>\na = 2(4) \u2013 3<br \/>\n= 8 \u2013 3 = 5<br \/>\nn<sup>th<\/sup> term = a + (n \u2013 1) d<br \/>\n\u2234 n<sup>th<\/sup> term = 5 + (n \u2013 1) 4<br \/>\n= 5 + 4n \u2013 4 = (4n + 1)<\/p>\n<p><strong>Question 56. The 16<sup>th<\/sup> term of an AP is 1 more than twice its 8<sup>th<\/sup> term. If the 12<sup>th<\/sup> term of the AP is 47, then find its n<sup>th<\/sup> term. (2012D)<\/strong><br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130873\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-19.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 19\" width=\"396\" height=\"172\" \/><br \/>\nFrom (i) and (ii), a = 4 \u2013 1 = 3<br \/>\nAs n<sup>th<\/sup> term = a + (n \u2013 1) d<br \/>\n\u2234 n<sup>th<\/sup> term = 3 + (n \u2013 1) 4<br \/>\n= 3 + 4n -4 = 4n -1<\/p>\n<p><strong>Question 57. A sum of \u20b91,600 is to be used to give ten cash prizes to students of a school for their overall academic performance. If each prize is \u20b920 less than its preceding prize, find the value of each of the prizes. (2012OD)<\/strong><br \/>\nSolution:<br \/>\nHere S<sub>10<\/sub> = 1600, d = -20, n = 10<br \/>\nS<sub>n<\/sub> = \\(\\frac{n}{2}\\) (2a + (n \u2013 1)d]\n\u2234 \\(\\frac{10}{2}\\)[2a + (10 \u2013 1)(-20)] = 1600<br \/>\n2a \u2013 180 = 320<br \/>\n2a = 320 + 180 = 500<br \/>\na = 250<br \/>\n\u2234 1<sup>st<\/sup> prize = a = \u20b9250<br \/>\n2<sup>nd<\/sup> prize = a<sub>2<\/sub> = a + d = 250 + (-20) = \u20b9230<br \/>\n3<sup>rd<\/sup> prize = a<sub>3<\/sub> = a<sub>2<\/sub> + d = 230 \u2013 20 = \u20b9210<br \/>\n4<sup>th<\/sup> prize = a<sub>4<\/sub> = a<sub>3<\/sub> + d = 210 \u2013 20 = \u20b9190<br \/>\n5<sup>th<\/sup> prize = a<sub>5<\/sub> = a<sub>4<\/sub> + d = 190 \u2013 20 = \u20b9170<br \/>\n6<sup>th<\/sup> prize = a<sub>6<\/sub> = a<sub>5<\/sub> + d = 170 \u2013 20 = \u20b9150<br \/>\n7<sup>th<\/sup> prize = a<sub>7<\/sub> = a<sub>6<\/sub> + d = 150 \u2013 20 = \u20b9130<br \/>\n8<sup>th<\/sup> prize = a<sub>8<\/sub> = a<sub>7<\/sub> + d = 130 \u2013 20 = \u20b9110<br \/>\n9<sup>th<\/sup> prize = a<sub>9<\/sub> = a<sub>8<\/sub> + d = 110 \u2013 20 = \u20b9590<br \/>\n10<sup>th<\/sup> prize = a<sub>10<\/sub> = a<sub>9<\/sub> + d = 90 \u2013 20 = \u20b970<br \/>\n= \u20b9 1,600<\/p>\n<p><strong>Question 58. Find the 60th term of the AP 8, 10, 12, \u2026, if it has a total of 60 terms and hence find the sum of its last 10 terms. (2015OD)<\/strong><br \/>\nSolution:<br \/>\na = 8, d = a<sub>2<\/sub> \u2013 a<sub>1<\/sub> = 10 \u2013 8 = 2, n = 60<br \/>\na<sub>60<\/sub> = a + 59d = 8 + 59(2) = 126<br \/>\n\u2234 Sum of its last 10 terms = S<sub>60<\/sub> \u2013 S<sub>50<\/sub><br \/>\n= \\(\\frac{n}{2}\\)(a + a<sub>n<\/sub>) \u2013 \\(\\frac{n}{2}\\)(2a + (n \u2013 1)d)<br \/>\n= \\(\\frac{60}{2}\\) (8 + a<sub>60<\/sub>) \u2013 \\(\\frac{50}{2}\\) (2 \u00d7 8 + (50 \u2013 1)2)<br \/>\n= 30 (8 + 126) \u2013 25 (16 + 98)<br \/>\n= 4020 \u2013 25 \u00d7 114<br \/>\n= 4020 \u2013 2850 = 1170<\/p>\n<p>Question 59.<br \/>\nAn Arithmetic Progression 5, 12, 19, \u2026 has 50 terms. Find its last term. Hence find the sum of its last 15 terms. (2015OD)<br \/>\nSolution:<br \/>\nLet a and d be the first term and common difference of A.P. respectively,<br \/>\na = 5, d = 12 \u2013 5 = 7, n = 50<br \/>\n\u2234 a<sub>n<\/sub> = a + (n \u2013 1)d<br \/>\na<sub>50<\/sub> = 5 + 49(7) = 5 + 343 = 348<br \/>\n\u2234 Sum of its last 15 terms = S<sub>50<\/sub> \u2013 S<sub>35<\/sub><br \/>\n= \\(\\frac{n}{2}\\)(a + a<sub>n<\/sub>) \u2013 \\(\\frac{n}{2}\\) (2a + (n \u2212 1)d)<br \/>\n= \\(\\frac{50}{2}\\) (5 + 348) \u2013 \\(\\frac{35}{2}\\) [2(5) + (35 \u2013 1)7]\n= 25(353) \u2013 \\(\\frac{35}{2}\\) (10 + 238)<br \/>\n= 8825 \u2013 35 \u00d7 124<br \/>\n= 8825 \u2013 4340 = 4485<\/p>\n<p>Question 60.<br \/>\nIf the sum of first 4 terms of an A.P. is 40 and that of first 14 terms is 280, find the sum of its first n terms. (2011D)<br \/>\nSolution:<br \/>\nS<sub>n<\/sub> = \\(\\frac{n}{2}\\)[2a + (n \u2212 1) d] \u2026(i)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130874\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-20.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 20\" width=\"398\" height=\"226\" \/><br \/>\nPutting the value of d = 2 in (ii), we get a = 7<br \/>\n\u2234 S<sub>n<\/sub> = \\(\\frac{n}{2}\\)[2(7) + (n \u2013 1). 2]\n= \\(\\frac{n}{2}\\).2 [7 + n \u2013 1]\n= n (n + 6) or n<sup>2<\/sup> + 6n<\/p>\n<p>Question 61.<br \/>\nThe first and the last terms of an A.P. are 8 and 350 respectively. If its common difference is 9, how many terms are there and what is their sum? (2011D)<br \/>\nSolution:<br \/>\nHere a = 8, a<sub>n<\/sub> = 350, d = 9<br \/>\nAs we know, a + (n \u2212 1) d = a<sub>2<\/sub><br \/>\n\u2234 8 + (n \u2013 1) 9 = 350<br \/>\n(n \u2212 1) 9 = 350 \u2013 8 = 342<br \/>\nn \u2013 1 = \\(\\frac{342}{9}\\) = 38<br \/>\nn = 38 + 1 = 39<br \/>\n\u2234 There are 39 terms.<br \/>\n\u2234 S<sub>n<\/sub> = \\(\\frac{n}{2}\\)(a + an)<br \/>\n\u2234 S<sub>39<\/sub> = \\(\\frac{39}{2}\\) (8 + 350) = \\(\\frac{39}{2}\\) \u00d7 358<br \/>\n= 39 \u00d7 179 = 6981<\/p>\n<p>Question 62.<br \/>\nSum of the first 20 terms of an AP is -240, and its first term is 7. Find its 24<sup>th<\/sup> term. (2012D)<br \/>\nSolution:<br \/>\nGiven: a = 7, S<sub>20<\/sub> = -240<br \/>\nHere, S<sub>n<\/sub> = \\(\\frac{n}{2}\\)[2a + (n \u2013 1)d]\n\u2234 S<sub>20<\/sub> = \\(\\frac{20}{2}\\)[2(7) + (20 \u2013 1)d]\n-240 = 10(14 + 19d)<br \/>\n\u2013\\(\\frac{240}{10}\\) = 14 + 19d = -24<br \/>\n19d = -24 \u2013 14 = -38<br \/>\n\u21d2 d = \\(\\frac{-38}{19}\\) = -2 \u2026(i)<br \/>\nAgain, a<sub>n<\/sub> = a + (n \u2013 1)d<br \/>\n\u2234 a<sub>24<\/sub> = 7 + (24 \u2013 1) (-2) \u2026 [From (i)<br \/>\n= 7 \u2013 46 = -39<\/p>\n<p>Question 63.<br \/>\nFind the common difference of an A.P. whose first term is 5 and the sum of its first four terms is half the sum of the next four terms. (2012OD)<br \/>\nSolution:<br \/>\nHere a = 5 \u2026 (i)<br \/>\nHere, a<sub>1<\/sub> + a<sub>2<\/sub> + a<sub>3<\/sub> + a<sub>4<\/sub> = \\(\\frac{1}{2}\\) (a<sub>5<\/sub> + a<sub>6<\/sub> + a<sub>7<\/sub> + a<sub>8<\/sub>)<br \/>\n\u2234 a + (a + d) + (a + 2d) + (a + 3d)<br \/>\n= \\(\\frac{1}{2}\\) [(a + 4d) + (a + 5d) + (a + 6d) + (a + 780] \u2026[a<sub>n<\/sub> = a(n \u2013 1)d<br \/>\n\u2234 4a + 6d = \\(\\frac{1}{2}\\) (4a + 22d)<br \/>\n8a + 12d = 4a + 22d<br \/>\n8a \u2013 4a = 22d \u2013 12d<br \/>\n4a = 100 \u21d2 4(5) = 10d<br \/>\nd = \\(\\frac{4(5)}{10}=\\frac{20}{10}\\) = 2 \u2026[From (1)<br \/>\n\u2234 Common difference, d = 2<\/p>\n<p>Question 64.<br \/>\nIf the sum of the first 7 terms of an A.P. is 119 and that of the first 17 terms is 714, find the sum of its first n terms. (2012OD)<br \/>\nSolution:<br \/>\nAs S<sub>n<\/sub> = \\(\\frac{n}{2}\\)[2a + (n \u2212 1)d]\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130875\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-21.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 21\" width=\"399\" height=\"275\" \/><br \/>\nS<sub>2<\/sub> = 119<br \/>\nFrom (i) and (ii), a = 17 \u2013 3(5) = 17 \u2013 15 = 2<br \/>\n\u2234S<sub>n<\/sub> = \\(\\frac{1}{2}\\)[2(2) + (n \u2013 1)5]\n= \\(\\frac{n}{2}\\)[4 + 5n \u2013 5]\n= \\(\\frac{n}{2}\\) (5n \u2013 1)<\/p>\n<p>Question 65.<br \/>\nThe 24<sup>th<\/sup> term of an AP is twice its 10th term. Show that its 72<sup>nd<\/sup> term is 4 times its 15th term. (2013D)<br \/>\nSolution:<br \/>\na<sub>24<\/sub> = 2 (10) \u2026 [Given<br \/>\na + 23d = 2 (a +9d) [\u2235 a<sub>n<\/sub> = a + (n \u2013 1)d)<br \/>\n23d = 2a + 18d \u2013 a<br \/>\n23d \u2013 18d = a \u21d2 a = 5d \u2026(i)<br \/>\nTo prove: a<sub>72<\/sub> = 4 (a<sub>15<\/sub>)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130876\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-22.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 22\" width=\"390\" height=\"162\" \/><br \/>\nFrom (ii) and (iii), L.H.S = R.H.S \u2026Hence Proved<\/p>\n<p>Question 66.<br \/>\nFind the number of terms of the A.P. -12, -9, \u2013 6, \u2026, 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained. (2013D)<br \/>\nSolution:<br \/>\nHere a = -12, d=-9 + 12 = 3<br \/>\nHere, a<sub>n<\/sub> = a + (n \u2013 1)d = 21<br \/>\n\u2234 -12 + (n \u2013 1).3 = 21<br \/>\n(n \u2013 1).3 = 21 + 12 = 33<br \/>\n\u2234 n \u2013 1 = 11<br \/>\nTotal number of terms,<br \/>\nn = 11 + 1 = 12<br \/>\nNew A.P. is<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130877\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-23.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 23\" width=\"314\" height=\"46\" \/><br \/>\nHere 1st term, a = -11<br \/>\nCommon difference, d = -8 + 11 = 3;<br \/>\nLast term, a<sub>n<\/sub> = 22; Number of terms, n= 12<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130878\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-24.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 24\" width=\"379\" height=\"152\" \/><\/p>\n<p>Question 67.<br \/>\nIn an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP. (2014D)<br \/>\nSolution:<br \/>\nHere, n = 50,<br \/>\nHere, S<sub>10<\/sub> = 210<br \/>\n= \\(\\frac{10}{2}\\) (2a + 9d) = 210 \u2026.[S<sub>n<\/sub> = \\(\\frac{1}{2}\\) [2a+(n \u2013 1)2]\n5(2a + 9d) = 210<br \/>\n2a + 9d = \\(\\frac{210}{5}\\) = 42<br \/>\n\u21d2 2a = 42 \u2013 9d \u21d2 a = \\(\\frac{42-9 d}{2}\\) \u2026(i)<br \/>\nNow, 50 = (1 + 2 + 3 + \u2026) + (36 + 37 + \u2026 + 50) Sum = 2565<br \/>\nSum of its last 15 terms = 2565 \u2026[Given<br \/>\nS<sub>50<\/sub> \u2013 S<sub>35<\/sub> = 2565<br \/>\n\\(\\frac{50}{2}\\)(2a + 49d) \u2013 \\(\\frac{35}{2}\\) (2a + 34d) = 2565<br \/>\n100a + 2450d \u2013 70a \u2013 1190d = 2565 \u00d7 2<br \/>\n30a + 1260d = 5130<br \/>\n3a + 1260 = 513 \u2026[Dividing both sides by 10<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130879\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-25.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 25\" width=\"394\" height=\"284\" \/><\/p>\n<p>Question 68.<br \/>\nIf the ratio of the sum of the first n terms of two A.Ps is (7n + 1): (4n + 27), then find the ratio of their 9<sup>th<\/sup> terms. (2017OD)<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130880\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-26.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 26\" width=\"393\" height=\"351\" \/><\/p>\n<p>Question 69.<br \/>\nIn a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two Sections, find how many trees were planted by the students. (2014OD)<br \/>\nSolution:<br \/>\nClasses: 1 + I + II + \u2026 + XII<br \/>\nSections: 2(I) + 2(II) + 2(III) + \u2026 + 2(XII)<br \/>\nTotal no. of trees<br \/>\n= 2 + 4 + 6 \u2026 + 24<br \/>\n= (2 \u00d7 2) + (2 \u00d7 4) + (2 \u00d7 6) + \u2026 + (2 \u00d7 24)<br \/>\n= 4 + 8 + 12 + \u2026 + 48<br \/>\n:: S<sub>12<\/sub> = \\(\\frac{12}{2}\\)(4 + 48) = 6(52) = 312 trees<\/p>\n<p>Question 70.<br \/>\nRamkali required \u20b9500 after 12 weeks to send her daughter to school. She saved \u20b9100 in the first week and increased her weekly saving by \u20b920 every week. Find whether she will be able to send her daughter to school after 12 weeks. (2015D)<br \/>\nSolution:<br \/>\nMoney required by Ramkali for admission of her daughter = \u20b92500<br \/>\nA.P. formed by saving<br \/>\n100, 120, 140, \u2026 upto 12 terms \u2026.(i)<br \/>\nLet, a, d and n be the first term, common difference and number of terms respectively. Here, a = 100, d = 20, n = 12<br \/>\nS<sub>n<\/sub> = \\(\\frac{n}{2}\\) (2a + (n \u2212 1)d)<br \/>\n\u21d2 S<sub>12<\/sub> = \\(\\frac{12}{2}\\) (2(100) + (12 \u2013 1)20)<br \/>\nS<sub>12<\/sub> = \\(\\frac{12}{2}\\) [2(100) + 11(20)] = 6[420] = \u20b92520<\/p>\n<p>Question 71.<br \/>\nThe sum of first n terms of an A.P. is 5n<sup>2<\/sup> + 3n. If its m<sup>th<\/sup> term is 168, find the value of m. Also find the 20<sup>th<\/sup> term of this A.P. (2013D)<br \/>\nSolution:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130881\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-27.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 27\" width=\"391\" height=\"280\" \/><\/p>\n<p>Question 72.<br \/>\nThe sum of first m terms of an AP is 4m<sup>2<\/sup> \u2013 m. If its n<sup>th<\/sup> term is 107, find the value of n. Also find the 21<sup>st<\/sup> term of this A.P. (2013D)<br \/>\nSolution:<br \/>\nWe have, S<sub>m<\/sub> = 4m<sup>2<\/sup> \u2013 m<br \/>\nPut m = 1,<br \/>\nS<sub>1<\/sub> = 4(1)<sup>2<\/sup> \u2013 (1)<br \/>\n= 4 \u2013 1 = 3<br \/>\nPut m = 2,<br \/>\nS<sub>2<\/sub> = 4(2)<sup>2<\/sup> \u2013 2<br \/>\n= 16 \u2013 2 = 14<br \/>\n\u2234 a<sub>1<\/sub> = S<sub>1<\/sub> = 3 = a<br \/>\na<sub>2<\/sub> = S<sub>2<\/sub> \u2013 S<sub>1<\/sub> = 14 \u2013 3 = 11<br \/>\nd = a<sub>2<\/sub> \u2013 a<sub>1<\/sub><br \/>\nd = 11 \u2013 3 = 8<br \/>\na<sub>n<\/sub> = a + (n \u2013 1)d = 107 \u2026[Given<br \/>\n\u2234 3+ (n \u2013 1)8 = 107<br \/>\n(n \u2013 1)8 = 107 \u2013 3 = 104<br \/>\n(n \u2013 1) = \\(\\frac{104}{8}\\) = 13<br \/>\nn = 13 + 1 = 14<br \/>\n\u2234 a<sub>21<\/sub> = a + 20d<br \/>\n= 3 + (20)8<br \/>\n= 3 + 160 = 163<\/p>\n<p>Question 73.<br \/>\nThe sum of first g terms of an A.P. is 63q \u2013 3q<sup>2<\/sup>. If its p<sup>th<\/sup> term is -60, find the value of p. Also find the 11<sup>th<\/sup> term of this A.P. (2013D)<br \/>\nSolution:<br \/>\nWe have, S<sub>q<\/sub> = 63q \u2013 3q<sup>2<\/sup><br \/>\nPut q = 1,<br \/>\nS<sub>7<\/sub> = 63(1) \u2013 3(1)<sup>2<\/sup><br \/>\n= 63 \u2013 3 = 60<br \/>\nPut q = 2,<br \/>\nS2 = 63(2) \u2013 3(2)<sup>2<\/sup><br \/>\n= 126 \u2013 12 = 114<br \/>\n\u2234 a = a<sub>1<\/sub> = S<sub>1<\/sub> = 60<br \/>\na<sub>2<\/sub> = S<sub>2<\/sub> \u2013 S7<sub>1<\/sub> = 114 \u2013 60 = 54<br \/>\nd = a<sub>2<\/sub> \u2013 a<sub>1<\/sub> = 54 \u2013 60 = -6<br \/>\np<sup>th<\/sup> term = -60<br \/>\na + (p \u2013 1)d = -60<br \/>\n60 + (p \u2013 1)(-6) = -60<br \/>\n(p \u2013 1)(-6) = -60 \u2013 60 = -120<br \/>\n(p \u2013 1) = \\(\\frac{-120}{-6}\\) = 20<br \/>\np= 20 + 1 = 21<br \/>\n\u2234 a<sub>11<\/sub> = a + 10d<br \/>\n= 60 + 10(-6)<br \/>\n= 60 \u2013 60 = 0<\/p>\n<p>Question 74.<br \/>\nFind the sum of the first 30 positive integers divisible by 6. (2011D)<br \/>\nSolution:<br \/>\nTo find: 6 + 12 + 18 + \u2026 (30 terms)<br \/>\nHere a = 6, d = 12 \u2013 6 = 6, n = 30<br \/>\nS<sub>n<\/sub> = [2a + (n \u2212 1) d]\n\u2234 S<sub>30<\/sub> = \\(\\frac{30}{2}\\)[2(6) + (30 \u2013 1) 6]\n= 15 [12 + 29(6)]\n= 15 (12 + 174)<br \/>\n= 15 (186) = 2790<\/p>\n<p>Question 75.<br \/>\nHow many multiples of 4 lie between 10 and 250? Also find their sum. (2011D)<br \/>\nSolution:<br \/>\nMultiples of 4 between 10 and 250 are:<br \/>\n12, 16, 20, \u2026\u2026\u2026 248<br \/>\nHere, a = 12, d = 4, a<sub>n<\/sub> = 248<br \/>\nAs we know, a + (n \u2013 1) d = a<sub>n<\/sub><br \/>\n\u223412 + (n \u2013 1) 4 = 248<br \/>\n\u21d2 (n \u2013 1) 4 = 248 \u2013 12 = 236<br \/>\nn \u2013 1 = \\(\\frac{236}{4}\\) = 59<br \/>\n\u21d2 n = 59 + 1 = 60<br \/>\n\u2234 There are 60 terms.<br \/>\nNow, S<sub>n<\/sub> = \\(\\frac{n}{2}\\) (a + a<sub>n<\/sub>)<br \/>\n\u2234 S<sub>60<\/sub> = \\(\\[\\frac{60}{2}\\)\\](12 + 248)<br \/>\n= 30 (260) = 7800<\/p>\n<p>Question 76.<br \/>\nFind the sum of all multiples of 8 lying between 201 and 950. (2012OD)<br \/>\nSolution:<br \/>\nTo find: 208 + 216 +224 + \u2026 + 944<br \/>\nHere, a = 208, d = 216 \u2013 208 = 8, a<sub>n<\/sub> = 944<br \/>\na + (n \u2013 1)d = a<sub>n<\/sub><br \/>\n\u2234 208 + (n \u2013 1)8 = 944<br \/>\n(n \u2013 1)8 = 944 \u2013 208 = 736<br \/>\nn \u2013 1 = \\(\\frac{736}{8}\\) = 92<br \/>\n8 n = 92 + 1 = 93<br \/>\nNow, S<sub>n<\/sub> = \\(\\frac{n}{2}\\)(a<sub>1<\/sub> + a<sub>n<\/sub>)<br \/>\n:: S<sub>93<\/sub> = \\(\\frac{93}{2}\\) (208 + 944)<br \/>\n= \\(\\frac{93}{2}\\) \u00d7 1152 = 93 \u00d7 576 = 53568<\/p>\n<p>Question 77.<br \/>\nFind the sum of all multiples of 9 lying between 400 and 800. (2012OD)<br \/>\nSolution:<br \/>\nTo find: 405 + 414 + 423 + \u2026 +792<br \/>\nHere a = 405, d = 414 \u2013 405 = 9, a<sub>n<\/sub> = 792<br \/>\na + (n \u2013 1)d = a<sub>n<\/sub><br \/>\n\u2234 405 + (n \u2013 1)9 = 792<br \/>\n(n \u2013 1)9 = 792 \u2013 405 = 387<br \/>\nn \u2013 1 = \\(\\frac{387}{9}\\) = 43<br \/>\n\u2234 n = 43 + 1 = 44<br \/>\nAs S<sub>n<\/sub> = \\(\\frac{n}{2}\\)(a<sub>1<\/sub> + a<sub>n<\/sub>)<br \/>\n\u2234 S<sub>44<\/sub> = \\(\\frac{44}{2}\\) (405 + 792)<br \/>\n= 22 \u00d7 1197 = 26334<\/p>\n<p>Question 78.<br \/>\nA thief runs with a uniform speed of 100 m\/minute. After one minute a policeman runs after the thief to catch him. He goes with a speed of 100 m\/minute in the first minute and increases his speed by 10 m\/minute every succeeding minute. After how many minutes the policeman will catch the thief. (2016D)<br \/>\nSolution:<br \/>\nLet total time ben minutes. Total distance covered by thief in n minutes<br \/>\n= Speed \u00d7 Time<br \/>\n= 100 \u00d7 n = 100 n metres<br \/>\nTotal distance covered by policeman<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130882\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-28.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 28\" width=\"396\" height=\"220\" \/><br \/>\n\u21d2 (n \u2013 1) (200 + 10n \u2013 20) = 200n<br \/>\n\u21d2 (n \u2013 1) [10n + 180) = 200n<br \/>\n\u21d2 10n<sup>2<\/sup> + 180n \u2013 10n \u2013 180 \u2013 200n = 0<br \/>\n\u21d2 10n<sup>2<\/sup> \u2013 30n \u2013 180 = 0<br \/>\n\u21d2 n<sup>2<\/sup> \u2013 3n \u2013 18 = 0 \u2026 [Dividing both sides by 10<br \/>\n\u21d2 n<sup>2<\/sup> \u2013 6n + 3n \u2013 18 = 0<br \/>\n\u21d2 n(n \u2013 6) + 3(n \u2013 6) = 0<br \/>\n\u21d2 (n + 3) (n \u2013 6) = 0<br \/>\n\u21d2 n + 3 = 0 or n \u2013 6 = 0<br \/>\n\u21d2 n = -3 or n = 6<br \/>\nBut n (time) can not be negative.<br \/>\n\u2234 Time taken by policeman to catch the thief<br \/>\n= n \u2013 1 = 6 \u2013 1 = 5 minutes<\/p>\n<p>Question 79.<br \/>\nA thief, after committing a theft, runs at a uniform speed of 50 m\/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m\/ minute every succeeding minute. After how many minutes, the policeman will catch the thief? (2016D)<br \/>\nSolution:<br \/>\nLet total time be n minutes.<br \/>\nTotal distance covered by thief in n minutes<br \/>\n= Speed \u00d7 Time<br \/>\n= (50 \u00d7 n) metres = 50 n metres<br \/>\nTotal distance covered by policeman<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130883\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-29.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 29\" width=\"393\" height=\"230\" \/><br \/>\n\u21d2 (n \u2013 2) (120 + 5n \u2013 15) = 100n<br \/>\n\u21d2 (n \u2013 2) [5n + 105) = 100n<br \/>\n\u21d2 5n<sup>2<\/sup> + 105n \u2013 10n \u2013 210 \u2013 100n = 0<br \/>\n\u21d2 5n<sup>2<\/sup> \u2013 5n \u2013 210 = 0<br \/>\n\u21d2 n<sup>2<\/sup> \u2013 n \u2013 42 = 0 \u2026[Dividing both sides by 5<br \/>\n\u21d2 n<sup>2<\/sup> \u2013 7n + 6n \u2013 42 = 0<br \/>\n\u21d2 n(n \u2013 7) + 6(n \u2013 7) = 0<br \/>\n\u21d2 (n \u2013 7) (n + 6) = 0<br \/>\n\u21d2 n \u2013 7 = 0 or n + 6 = 0<br \/>\n\u21d2 n = 7 or n = -6 (reject)<br \/>\nBut n (time) can not be negative.<br \/>\n\u2234 Time taken by policeman to catch the thief<br \/>\n= n \u2013 2 = 7 \u2013 2 = 5 minutes<\/p>\n<p>Question 80.<br \/>\nThe houses in a row are numbered conse cutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X. (2016OD)<br \/>\nSolution:<br \/>\nHere the A.P. is 1, 2, 3, \u2026., 49<br \/>\nHere a = 1, d = 1, a<sub>n<\/sub> = 49<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130884\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-30.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 30\" width=\"387\" height=\"136\" \/><br \/>\nNow,<br \/>\n\u21d2 (X \u2013 1)<sup>2<\/sup> (2 + (X \u2013 2)) = 49(2 + 48) \u2013 X[2 + (x \u2013 1)]\n\u21d2 (X \u2013 1). X = 2,450 \u2013 X(X + 1)<br \/>\n\u21d2 x<sup>2<\/sup> \u2013 X = 2,450 \u2013 X<sup>2<\/sup> \u2013 X<br \/>\n\u21d2 X<sup>2<\/sup> \u2013 X + X<sup>2<\/sup> + X = 2,450<br \/>\n\u21d2 2X<sup>2<\/sup> = 2,450<br \/>\n\u21d2 X<sup>2<\/sup> = 1,225<br \/>\n\u2234 X = \\(+\\sqrt{1,225}\\) = 35 \u2026[X can not be -ve<\/p>\n<p>Question 81.<br \/>\nFind the sum of n terms of the series \\(\\left(4-\\frac{1}{n}\\right)+\\left(4-\\frac{2}{n}\\right)+\\left(4-\\frac{3}{n}\\right)+\\ldots \\ldots .\\) (2017D)<br \/>\nSolution:<br \/>\nFirst term, a = 4 \u2013 \\(\\frac{1}{n}\\)<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-130885\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2021\/12\/Important-Questions-for-Class-10-Maths-Chapter-5-Arithmetic-Progressions-31.png\" alt=\"Important Questions for Class 10 Maths Chapter 5 Arithmetic Progressions 31\" width=\"334\" height=\"328\" \/><\/p>\n<h4><\/h4>\n","protected":false},"excerpt":{"rendered":"<p>Arithmetic Progressions Class 10 Important Questions Very Short Answer (1 Mark) Question 1. 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