{"id":3202,"date":"2021-12-29T12:40:29","date_gmt":"2021-12-29T07:10:29","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=3202"},"modified":"2024-07-03T14:47:00","modified_gmt":"2024-07-03T09:17:00","slug":"ncert-solutions-for-class-11-maths-chapter-9-sequence-and-series-exercise-9-3","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-solutions\/class-11\/maths\/chapter-9-sequence-and-series-exercise-9-3\/","title":{"rendered":"NCERT Solutions for Class 11 Maths Chapter 9 &#8211; Sequences and Series Exercise 9.3"},"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-materials\/ncert-solutions\/class-11\/maths\/chapter-9-sequence-and-series-exercise-9-3\/#NCERT_Solutions_for_Class_11_Maths_Chapter_9_-_Sequences_and_Series_Exercise_93_-_PDF_Download\" title=\"NCERT Solutions for Class 11 Maths Chapter 9 &#8211; Sequences and Series Exercise 9.3 &#8211; PDF Download\">NCERT Solutions for Class 11 Maths Chapter 9 &#8211; Sequences and Series Exercise 9.3 &#8211; PDF Download<\/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-materials\/ncert-solutions\/class-11\/maths\/chapter-9-sequence-and-series-exercise-9-3\/#NCERT_Solutions_for_Class_11_Maths_Chapter_9_Sequence_and_series_Exercise_93\" title=\"NCERT Solutions for Class 11 Maths Chapter 9 Sequence and series Exercise 9.3\">NCERT Solutions for Class 11 Maths Chapter 9 Sequence and series Exercise 9.3<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<p>Chapter 9: Sequences and Series in Class 11 Maths is part of the <a href=\"https:\/\/infinitylearn.com\/surge\/cbse\/cbse-syllabus\/\"><strong>CBSE Syllabus<\/strong><\/a> for 2024-25. The <a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-solutions\/\"><strong>NCERT Solutions<\/strong><\/a> for this chapter help students master problems related to sequences and series. Subject experts have solved all the questions in this exercise. Exercise 9.3 of <a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-solutions\/class-11\/maths\/\"><strong>NCERT Solutions for Class 11 Maths<\/strong><\/a> Chapter 9 covers:<\/p>\n<ul>\n<li>Geometric Progression (G.P.)<\/li>\n<li>The general term of a G.P.<\/li>\n<li>Sum to n terms of a G.P.<\/li>\n<li>Geometric Mean (G.M.)<\/li>\n<li>Relationship between A.M. and G.M.<\/li>\n<\/ul>\n<p>These solutions are prepared by subject matter experts at Infinity Learn, providing detailed methods for solving problems. By understanding the concepts in these solutions, students can clear their doubts and excel in their board exams.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_11_Maths_Chapter_9_-_Sequences_and_Series_Exercise_93_-_PDF_Download\"><\/span>NCERT Solutions for Class 11 Maths Chapter 9 &#8211; Sequences and Series Exercise 9.3 &#8211; PDF Download<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\t\t<div class=\"download-pdf-container\">\n\t\t\t<span class=\"title\">Do you need help with your Homework? 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href=\"\/surge\/?pdfemb-serveurl=https%3A%2F%2Finfinitylearn.com%2Fsurge%2Fwp-content%2Fuploads%2Fsecurepdfs%2F2024%2F07%2FNCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Exercise-9.3.pdf\" class=\"pdfemb-viewer\" style=\"\" data-width=\"max\" data-height=\"max\" data-mobile-width=\"0\"  data-scrollbar=\"both\" data-download=\"off\" data-tracking=\"on\" data-newwindow=\"on\" data-pagetextbox=\"off\" data-scrolltotop=\"on\" data-startzoom=\"100\" data-startfpzoom=\"100\" data-download-nonce=\"0da46c77df\" data-disablerightclick=\"on\" data-toolbar=\"top\" data-toolbar-fixed=\"on\">NCERT-Solutions-for-Class-11-Maths-Chapter-9-Sequences-and-Series-Exercise-9.3<br\/><\/a>\n<h3><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_11_Maths_Chapter_9_Sequence_and_series_Exercise_93\"><\/span>NCERT Solutions for Class 11 Maths Chapter 9 Sequence and series Exercise 9.3<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. The 4<sup>th<\/sup> term of a G.P. is the square of its second term, and the first term is \u20133. Determine its 7<sup>th<\/sup> term.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s consider <em>a<\/em> to be the first term and <em>r<\/em> to be the common ratio of the G.P.<\/p>\n<p>Given, <em>a<\/em> = \u20133<\/p>\n<p>And we know that,<\/p>\n<p><em>a<sub>n<\/sub><\/em> = <em>ar<sup>n<\/sup><\/em><sup>\u20131<\/sup><\/p>\n<p>So, <em>a<\/em><sub>4 <\/sub>= <em>ar<\/em><sup>3<\/sup> = (\u20133) <em>r<\/em><sup>3<\/sup><\/p>\n<p><em>a<\/em><sub>2<\/sub> = <em>a r<\/em><sup>1<\/sup> = (\u20133) <em>r<\/em><\/p>\n<p>Then from the question, we have<\/p>\n<p>(\u20133) <em>r<\/em><sup>3<\/sup> = [(\u20133) <em>r<\/em>]<sup>2<\/sup><\/p>\n<p>\u21d2 \u20133<em>r<\/em><sup>3<\/sup> = 9 <em>r<\/em><sup>2<\/sup><\/p>\n<p>\u21d2 <em>r<\/em> = \u20133<\/p>\n<p><em>a<\/em><sub>7<\/sub> = <em>a<\/em> <em>r<\/em> <sup>7\u20131 <\/sup>= <em>a<\/em> <em>r<\/em><sup>6<\/sup> = (\u20133) (\u20133)<sup>6<\/sup> = \u2013 (3)<sup>7<\/sup> = \u20132187<\/p>\n<p>Therefore, the seventh term of the G.P. is \u20132187.<\/p>\n<p><strong>2. The sum of the first three terms of a G.P. is 39\/10, and their product is 1. Find the common ratio and the terms.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let a\/r, a, ar be the first three terms of the G.P.<\/p>\n<p>a\/r + a + ar = 39\/10 \u2026\u2026 (1)<\/p>\n<p>(a\/r) (a) (ar) = 1 \u2026\u2026.. (2)<\/p>\n<p>From (2), we have<\/p>\n<p>a<sup>3<\/sup> = 1<\/p>\n<p>Hence, a = 1 [Considering real roots only]\n<p>Substituting the value of a in (1), we get<\/p>\n<p>1\/r + 1 + r = 39\/10<\/p>\n<p>(1 + r + r<sup>2<\/sup>)\/r = 39\/10<\/p>\n<p>10 + 10r + 10r<sup>2<\/sup> = 39r<\/p>\n<p>10r<sup>2<\/sup> \u2013 29r + 10 = 0<\/p>\n<p>10r<sup>2<\/sup> \u2013 25r \u2013 4r + 10 = 0<\/p>\n<p>5r(2r \u2013 5) \u2013 2(2r \u2013 5) = 0<\/p>\n<p>(5r \u2013 2) (2r \u2013 5) = 0<\/p>\n<p>Thus,<\/p>\n<p>r = 2\/5 or 5\/2<\/p>\n<p>Therefore, the three terms of the G.P. are 5\/2, 1 and 2\/5.<\/p>\n<p><strong>3. If the p<sup>th<\/sup>, q<sup>th<\/sup> and r<sup>th<\/sup> terms of a G.P. are <em>a, b<\/em> and <em>c<\/em>, respectively. Prove that a<sup>q-r <\/sup>b<sup>r-p <\/sup>c<sup>p-q<\/sup> = 1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s take <em>A<\/em> to be the first term and <em>R<\/em> to be the common ratio of the G.P.<\/p>\n<p>Then according to the question, we have<\/p>\n<p><em>AR<sup>p<\/sup><\/em><sup>\u20131 <\/sup>= <em>a<\/em><\/p>\n<p><em>AR<sup>q<\/sup><\/em><sup>\u20131 <\/sup>= <em>b<\/em><\/p>\n<p><em>AR<sup>r<\/sup><\/em><sup>\u20131 <\/sup>= <em>c<\/em><\/p>\n<p>Then,<\/p>\n<p><em>a<sup>q\u2013r<\/sup><\/em><em>b<sup>r\u2013p<\/sup><\/em><em>c<sup>p\u2013q<\/sup><\/em><\/p>\n<p>= <em>A<sup>q<\/sup><\/em><sup>\u2013<\/sup><em><sup>r <\/sup><\/em>\u00d7 <em>R<\/em><sup>(<\/sup><em><sup>p<\/sup><\/em><sup>\u20131) (q\u2013r)<\/sup> \u00d7 A<em><sup>r<\/sup><\/em><sup>\u2013<\/sup><em><sup>p<\/sup><\/em> \u00d7 <em>R<\/em><sup>(<\/sup><em><sup>q<\/sup><\/em><sup>\u20131) (<\/sup><em><sup>r<\/sup><\/em><sup>\u2013<\/sup><em><sup>p<\/sup><\/em><sup>)<\/sup> \u00d7 <em>A<sup>p<\/sup><\/em><sup>\u2013<\/sup><em><sup>q<\/sup><\/em> \u00d7 <em>R<\/em><sup>(<\/sup><em><sup>r <\/sup><\/em><sup>\u20131)(<\/sup><em><sup>p<\/sup><\/em><sup>\u2013<\/sup><em><sup>q<\/sup><\/em><sup>)<\/sup><\/p>\n<p>= <em>Aq<\/em><sup> \u2013 <\/sup><em><sup>r<\/sup><\/em><sup> + <\/sup><em><sup>r<\/sup><\/em><sup> \u2013 <\/sup><em><sup>p<\/sup><\/em><sup> + <\/sup><em><sup>p<\/sup><\/em><sup> \u2013 <\/sup><em><sup>q<\/sup><\/em> \u00d7 <em>R<\/em> <sup>(<\/sup><em><sup>pr<\/sup><\/em><sup> \u2013 <\/sup><em><sup>pr<\/sup><\/em><sup> \u2013 <\/sup><em><sup>q<\/sup><\/em><sup> + <\/sup><em><sup>r<\/sup><\/em><sup>) + (<\/sup><em><sup>rq<\/sup><\/em><sup> \u2013<\/sup><em><sup> r <\/sup><\/em><sup>+ <\/sup><em><sup>p<\/sup><\/em><sup> \u2013 <\/sup><em><sup>pq<\/sup><\/em><sup>) + (<\/sup><em><sup>pr<\/sup><\/em><sup> \u2013 <\/sup><em><sup>p<\/sup><\/em><sup> \u2013 <\/sup><em><sup>qr<\/sup><\/em><sup> + <\/sup><em><sup>q<\/sup><\/em><sup>)<\/sup><\/p>\n<p>= <em>A<\/em><sup>0<\/sup> \u00d7 <em>R<\/em><sup>0<\/sup><\/p>\n<p>= 1<\/p>\n<p>Hence proved.<\/p>\n<p><strong>4. If <em>a, b, c<\/em> and <em>d<\/em> are in G.P., show that (a<sup>2<\/sup> + b<sup>2<\/sup> + c<sup>2<\/sup>)(b<sup>2<\/sup> + c<sup>2<\/sup> + d<sup>2<\/sup>) = (ab + bc + cd)<sup>2<\/sup>.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, <em>a<\/em>, <em>b<\/em>, <em>c<\/em>, <em>d<\/em> are in G.P.<\/p>\n<p>So, we have<\/p>\n<p><em>bc<\/em> = <em>ad<\/em> \u2026 (1)<\/p>\n<p><em>b<\/em><sup>2<\/sup> = <em>ac <\/em>\u2026 (2)<\/p>\n<p><em>c<\/em><sup>2<\/sup> = <em>bd<\/em> \u2026 (3)<\/p>\n<p>Taking the R.H.S., we have<\/p>\n<p>R.H.S.<\/p>\n<p>= (<em>ab<\/em> + <em>bc<\/em> + <em>cd<\/em>)<sup>2<\/sup><\/p>\n<p>= (<em>ab<\/em> + <em>ad <\/em>+ <em>cd<\/em>)<sup>2<\/sup> [Using (1)]\n<p>= [<em>ab<\/em> + <em>d<\/em> (<em>a<\/em> + <em>c<\/em>)]<sup>2<\/sup><\/p>\n<p>= <em>a<\/em><sup>2<\/sup><em>b<\/em><sup>2<\/sup> + 2<em>abd<\/em> (<em>a<\/em> + <em>c<\/em>) + <em>d<\/em><sup>2<\/sup> (<em>a<\/em> + <em>c<\/em>)<sup>2<\/sup><\/p>\n<p>= <em>a<\/em><sup>2<\/sup><em>b<\/em><sup>2<\/sup> +2<em>a<\/em><sup>2<\/sup><em>bd<\/em> + 2<em>acbd<\/em> + <em>d<\/em><sup>2<\/sup>(<em>a<\/em><sup>2<\/sup> + 2<em>ac<\/em> + <em>c<\/em><sup>2<\/sup>)<\/p>\n<p>= <em>a<\/em><sup>2<\/sup><em>b<\/em><sup>2<\/sup> + 2<em>a<\/em><sup>2<\/sup><em>c<\/em><sup>2<\/sup> + 2<em>b<\/em><sup>2<\/sup><em>c<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup><em>a<\/em><sup>2<\/sup> + 2<em>d<\/em><sup>2<\/sup><em>b<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup><em>c<\/em><sup>2<\/sup> [Using (1) and (2)]\n<p>= <em>a<\/em><sup>2<\/sup><em>b<\/em><sup>2<\/sup> + <em>a<\/em><sup>2<\/sup><em>c<\/em><sup>2<\/sup> + <em>a<\/em><sup>2<\/sup><em>c<\/em><sup>2<\/sup> + <em>b<\/em><sup>2<\/sup><em>c<\/em><sup>2 <\/sup>+ <em>b<\/em><sup>2<\/sup><em>c<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup><em>a<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup><em>b<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup><em>b<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup><em>c<\/em><sup>2<\/sup><\/p>\n<p>= <em>a<\/em><sup>2<\/sup><em>b<\/em><sup>2<\/sup> + <em>a<\/em><sup>2<\/sup><em>c<\/em><sup>2<\/sup> + <em>a<\/em><sup>2<\/sup><em>d<\/em><sup>2 <\/sup>+ <em>b<\/em><sup>2 <\/sup>\u00d7 <em>b<\/em><sup>2<\/sup> + <em>b<\/em><sup>2<\/sup><em>c<\/em><sup>2<\/sup> + <em>b<\/em><sup>2<\/sup><em>d<\/em><sup>2<\/sup> + <em>c<\/em><sup>2<\/sup><em>b<\/em><sup>2<\/sup> + <em>c<\/em><sup>2 <\/sup>\u00d7 <em>c<\/em><sup>2<\/sup> + <em>c<\/em><sup>2<\/sup><em>d<\/em><sup>2<\/sup><\/p>\n[Using (2) and (3) and rearranging terms]\n<p>= <em>a<\/em><sup>2<\/sup>(<em>b<\/em><sup>2<\/sup> + <em>c<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup>) + <em>b<\/em><sup>2<\/sup> (<em>b<\/em><sup>2<\/sup> + <em>c<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup>) + <em>c<\/em><sup>2<\/sup> (<em>b<\/em><sup>2<\/sup>+ <em>c<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup>)<\/p>\n<p>= (<em>a<\/em><sup>2<\/sup> + <em>b<\/em><sup>2<\/sup> + <em>c<\/em><sup>2<\/sup>) (<em>b<\/em><sup>2<\/sup> + <em>c<\/em><sup>2<\/sup> + <em>d<\/em><sup>2<\/sup>)<\/p>\n<p>= L.H.S.<\/p>\n<p>Thus, L.H.S. = R.H.S.<\/p>\n<p>Therefore,<\/p>\n<p>(a<sup>2<\/sup> + b<sup>2<\/sup> + c<sup>2<\/sup>)(b<sup>2<\/sup> + c<sup>2<\/sup> + d<sup>2<\/sup>) = (ab + bc + cd)<sup>2<\/sup><\/p>\n<p><strong>5. Insert two numbers between 3 and 81 so that the resulting sequence is G.P.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume <em>G<\/em><sub>1<\/sub> and <em>G<\/em><sub>2<\/sub> to be two numbers between 3 and 81 such that the series 3, <em>G<\/em><sub>1<\/sub>, <em>G<\/em><sub>2<\/sub>, 81 forms a G.P.<\/p>\n<p>And let <em>a<\/em> be the first term and <em>r<\/em> be the common ratio of the G.P.<\/p>\n<p>Now, we have the 1<sup>st<\/sup> term as 3 and the 4<sup>th<\/sup> term as 81.<\/p>\n<p>81 = (3) <em>(r<\/em>)<sup>3<\/sup><\/p>\n<p><em>r<\/em><sup>3<\/sup> = 27<\/p>\n<p>\u2234 <em>r<\/em> = 3 (Taking real roots only)<\/p>\n<p>For <em>r<\/em> = 3,<\/p>\n<p><em>G<\/em><sub>1<\/sub> = <em>ar<\/em> = (3) (3) = 9<\/p>\n<p><em>G<\/em><sub>2<\/sub> = <em>ar<\/em><sup>2<\/sup> = (3) (3)<sup>2<\/sup> = 27<\/p>\n<p>Therefore, the two numbers which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P. are 9 and 27.<\/p>\n<p><strong>6. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2<sup>nd<\/sup> hour, 4<sup>th<\/sup> hour and <em>n<\/em><sup>th<\/sup> hour?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given, the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.<\/p>\n<p>Here, we have <em>a<\/em> = 30 and <em>r<\/em> = 2<\/p>\n<p>So, <em>a<\/em><sub>3<\/sub> = <em>ar<\/em><sup>2<\/sup> = (30) (2)<sup>2<\/sup> = 120<\/p>\n<p>Thus, the number of bacteria at the end of 2<sup>nd<\/sup> hour will be 120.<\/p>\n<p>And, <em>a<\/em><sub>5<\/sub> = <em>ar<\/em><sup>4<\/sup> = (30) (2)<sup>4<\/sup> = 480<\/p>\n<p>The number of bacteria at the end of 4<sup>th<\/sup> hour will be 480.<\/p>\n<p><em>a<sub>n<\/sub><\/em><sub> +1 <\/sub>= <em>ar<sup>n<\/sup><\/em> = (30) 2<em><sup>n<\/sup><\/em><\/p>\n<p>Therefore, the number of bacteria at the end of <em>n<\/em><sup>th<\/sup> hour will be 30(2)<em><sup>n<\/sup><\/em>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Chapter 9: Sequences and Series in Class 11 Maths is part of the CBSE Syllabus for 2024-25. 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