NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions – Infinity Learn

For the advantage of students studying for the CBSE Term II Class 10 Maths test, NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression are offered here. Students must become familiar with these NCERT Exemplar Solutions to achieve a good grade in the Class 10 examination. The specialists at Infinity Learn answer these NCERT difficulties. These answers will assist you in comprehending and mastering various types of arithmetic progressions questions. NCERT Exemplar Solutions assist you in achieving proficiency in solving all types of queries.

NCERT Exemplar Class 10 Maths Chapter 5 Exercise 5.1

Choose the correct answers from the given four options:
Question 1.
In an AP, if d = -4, n = 7, a_n = 4, then a is equal to
(A) 6
(B) 7
(C) 20
(D) 28
Solution:
(D)
In an AP,
a_n = a + (n – 1 )d
⇒ 4 = a + (7 – 1) (-4) [by given conditions] ⇒ 4 = a + 6(-4) ⇒ 4 + 24 = a
∴ a = 28

Question 2.
In an AP, if a = 3.5, d = 0,n = 101, then a_n will be
(A) 0
(B) 3.5
(C) 103.5
(D) 104.5
Solution:
(B)
In an AP,
a_n = a + (n – 1)d = 3.5 + (101 – 1) × 0
[by given conditions] ∴. a_n = 3.5

Question 3.
The list of numbers – 10, – 6, – 2, 2,… is
(A) an AP with d = -16
(B) anAP with d = 4
(C) an AP with d = – 4
(D) not an AP
Solution:
(B)
The given numbers are -10, -6, -2, 2,
Here, a₁ = -10, a₂ = -6, a₃ = 2 and a₄ = 2,….
Since a₂ – a₁ = -6 – (-10) = -6 + 10 = 4
a₃ – a₂ = -2 – (-6) = -2 + 6 = 4
a₄ – a₃ = 2-(-2) = 2 + 2 = 4
………….
…………
………….
Each successive term of given list has same difference i.e., 4
So, the given list forms an AP with common difference, d = 4

Question 4.
The 11th term of the AP: -5, −52 , 0, 52,…
(A) -20
(B) 20
(C) -30
(D) 30
Solution:
(B)
Given AP -5, −52 , 0, 52,…
Here, a = -5, d = −52+5=52
∴ a_n = a + (11 – l)d [∵ a_n = a + (n – 1)d] = -5 + (10) × 52
= -5 + 25 = 20

Question 5.
The first four terms of an AP, whose first term is -2 and the common difference is -2, are
(A) -2, 0, 2, 4
(B) -2, 4, -8, 16
(C) -2, -4, -6, -8
(D) -2, -4, -8, -16
Solution:
(C)
Let the first four terms of an AP are
a, a + d, a + 2d and a + 3d
Given, that first term, a = -2 and common
difference, d = -2, then we have an AP as follows
-2, -2 -2, -2 + 2(-2), -2 + 3(-2)
= -2, -4, -6, -8

Question 6.
The 21st term of the AP whose first two terms are -3 and 4 is
(A) 17
(B) 137
(C) 143
(D) -143
Solution:
(B)
Given, first two terms of an AP are a = -3 and a + d = 4
⇒ — 3 + d = 4
Common difference, d = 7
⇒ a₂₁ = a + (21 – 1)d [∵ a_n = a + (n – 1)d] = -3 +(20)7
= -3 + 140 = 137

Question 7.
If the 2nd term ofanAPis13and the 5th term is 25, what is its 7th term?
(A) 30
(B) 33
(C) 37
(D) 38
Solution:
(B)
Given a₂ = 13 and a₅ = 25
⇒ a + (2 – 1)d = 13 [∵ a_n = a + (n – 1)d] and a + (5 – 1 )d = 25
⇒ a + d = 13 ……….. (i)
and a + 4d = 25 ………… (ii)
On subtracting Eq. (i) from Eq. (ii), we get
3d = 25 – 13 = 12
⇒ d = 4
From Eq. (i) a = 13 – 4 = 9
∴ a₇ = a + (7 – 1)d = 9 + 6 × 4 = 33

Question 8.
Which term of the AP: 21, 42, 63, 84,… is 210?
(A) 9^th
(B) 10^th
(C) 11^th
(D) 12^th
Solution:
(B)
Let n^th term of the given AP be 210.
Here, first term, a = 21
and common difference, d = 42 – 21 = 21 and a_n = 210
∵ a_n = a + (n – 1 )d
⇒ 210 = 21 + (n- 1)21
⇒ 210 = 21 + 21n- 21
⇒ 210 = 21n ⇒ n = 10
Hence, the 10th term of an AP is 210.

Question 9.
If the common difference of an AP is 5, then what is a₁₈ – a₁₃?
(A) 5
(B) 20
(C) 25
(D) 30
Solution:
(C)
Given, the common difference of AP i.e., d = 5
Now, a₁₈ – a₁₃ = a + (18 – 1)d – [a + (13 – 1)d] [∵ a_n = a + (n – 1)d] = a + (17 × 5) – a – (12 × 5)
= 85 – 60 = 25

Question 10.
What is the common difference of an AP in which a₁₈ – a₁₄ = 32?
(A) 8
(B) -8
(C) -4
(D) 4
Solution
(A)
Given, a₁₈ – a₁₄ = 32
⇒ a + (18 – 1)d – [a + (14 – 1)d] = 32
[∵a_n = a + (n – 1 )d] ⇒ a + 17d – a – 13d = 32
⇒ 4d = 32
∴ d = 8
which is the required common difference of an AP.

Question 11.
Two APs have the same common difference. The first term of one of these is -1 and that of the other is – 8. Then the difference between their 4th terms is
(A) -1
(B) -8
(C) 7
(D) -9
Solution:
(C)
Let the common difference of two APs are d₁ and d₂, respectively.
By given condition, d₁ = d₂ = d ……….. (i)
Let the first term of first AP(a₁) = -1
and the first term of second AP (a₂) = -8
We know that, the nth term of an AP
T_n = a + (n – 1)d
∴ 4th term of first AP,
T₄ = a₁ + (4 – 1)d = -1 + 3d
and 4th term of second AP,
T₄‘ = a₂ + (4 – 1)d = -8 + 3d
Now, the difference between their 4th terms is i.e.,
∣∣T4−T′4∣∣ = (-1 + 3d) – (-8 + 3d)
= -1 + 3d + 8 – 3d = 7
Hence, the required difference is 7.

Question 12.
If 7 times the 7^th term of an AP is equal to 11 times its 11th term, then its 18^th term will be
(A) 7
(B) 11
(C) 18
(D) 0
Solution:
(D)
According to the question,
7a₇ = 11a₁₁
⇒ 7[a + (7 – 1)d] = 11 [d + (11 – 1)d] [∵ a_n = a + (n – 1 )d] ⇒ 7(o + 6d = 11 (a + 10d)
⇒ 7a + 42d = 11a + 10d
⇒ 4a + 68d = 0
⇒ 2(2a + 34d) = 0
⇒ 2a + 34d = 0
⇒ a + 17d = 0 …………. (i)
∴ 18^th term of an AP, a₁₈ = a + (18 – 1)d = a + 17d = 0
[from Eq(i)]

Question 13.
The 4th term from the end of the AP: -11, -8, -5, 49 is
(A) 37
(B) 40
(C) 43
(D) 58
Solution:
(B)
We know that, the nth term of an AP from the end is
a_n = l -(n – 1 )d …………. (i)
Here, l = Last term = 49 [given] Common difference, d = -8 – (-11) = -8 + 11 = 3
From Eq (i), a₄ = 49 – (4 – 1) 3 = 49 – 9 = 40

Question 14.
The famous mathematician associated with finding the sum of the first 100 natural numbers is
(A) Pythagoras
(B) Newton
(C) Gauss
(D) Euclid
Solution:
(C)
Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers i.e., 1, 2, 3,100.

Question 15.
If the first term of an AP is -5 and the common difference is 2, then the sum of the first 6 terms is
(A) 0
(B) 5
(C) 6
(D) 15
Solution:
(A)
Given, a = -5 and d = 2
∴ S₆ = 62 [2a + (6 – 1)d] [∵ S_n = n2 [2a + (n – 1)d] = 3[2(-5) + 5(2)] = 3(-10 + 10) = 0

Question 16.
The sum of first 16 terms of the AP: 10, 6 2, … is
(A) -320
(B) 320
(C) -352
(D) -400
Solution:
(A)
Given, AP is 10, 6, 2,
Here, first term a = 10, common difference, d = – 4
∴ S₁₆ = 162 [2a + (16 – 1)d] [∵ S_n = n2 [2a + (n – 1)d] = 8[2 × 10 + 15(-4)] = 8(20 – 60) = 8(-40) = -320

Question 17.
In an AP if a = 1, a„ = 20 and S_n = 399, then n is
(A) 19
(B) 21
(C) 38
(D) 42
Solution:
(C)
[∵ S_n = n2 [2a + (n – 1)d] 399 = n2 [2 × 1 + (n – 1)d] [∵ a = 1] 798 = 2n + n(n – 1)d ………… (i)
and, a_n = 20
⇒ a + (n-l)d = 20 [∵ a_n = a + (n – 1)d] ⇒ 1 + (n – 1)d = 20⇒ (n – 1d = 19 ……… (ii)
Using Eq. (ii) in Eq. (i), we get
798 = 2n + 19 n
⇒ 798 = 21n
∴ n = 79821

Question 18.
The sum of first five multiples of 3 is
(A) 45
(B) 55
(C) 65
(D) 75
Solution:
(A)
The first five multiples of 3 are 3, 6, 9,12 and 15.
Here, first term, a = 3, common difference, d = 6 – 3 = 3 and number of terms n = 5
∴ S₅ = 52 [2a + (5 – 1)d] [∵ S_n = n2 [2a + (n – 1)d] = 52 [2 × 3 + 4 × 3] = 52 (6 + 12) = 5 × 9 = 45

NCERT Exemplar Class 10 Maths Chapter 5 Exercise – 5.2

Question 1.
Which of the following form an AP? Justify your answer.
(i) -1,-1,-1,-1,…
(ii) 0,2,0,2,…
(iii) 1,1,2,2,3,3,…
(iv) 11,22,33,…
(v) 12,13,14,….
(vi) 2,2², 2³, 2⁴, …
(vii) 3–√,12−−√,27−−√,48−−√,…
Solution:
(i) Here, t₁ = -1, t₂ = -1, t₃ = -1 and t₂ = -1
t₂ – t₁ = -1 + 1 = 0
t₃ – t₂ = —1 + 1 = 0
t₄ – t₃ = -1 + 1 = 0
Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

(ii) Here, t₁ = 0, t₂ = 2, t₃ = 0 and t₄ = 2
t₂ – t₁ = 2 – 0 = 2
t₃ – t₂ = 0 – 2 = -2
t₄ – t₃ = 2 – 0 = 2
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iii) Here, t₁ = 1, t₂ = 1, t₃ = 2 and t₄ = 2
t₂ – t₁ = 1 -1 = 0
t₃ – t₂ = 2 – 1 = 1
t₄ – t₃ = 2 – 2 = 0
Clearly, the difference of successive terms is not same, therefore given list of numbers does not form an AP.

(iv) Here, t₁= 11, t₂ = 22 and t₃ = 33
t₂ – t₁ = 22 – 11 = 11
t₃ – t₂ = 33 – 22 = 11
Clearly, the difference of successive terms is same, therefore given list of numbers form an AP.

Question 3.
For the AP: -3, -7, -11,can we find directly a₃₀ – a₂₀ without actually finding a₃₀ and a₂₀? Give reasons for your answer.
Solution:
True
∵ nth term of an AP, a_n = a + (n – 1)d
∴ a₃₀ = a + (30 – 1 )d = a + 29d
and a₂₀ = a + (20 -1)d = a + 19d ………… (i)
Now, a₃₀ – a₂₀ = -(a + 29d) – (a + 19d) = 10d
and from given AP,
Common difference, d = -7 – (-3) = -7 + 3 = -4
∴ a₃₀ – a₂₀ = 10(-4) = -40 [From (i)]

Question 4.
Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10^th terms is the same as the difference between their 21^st terms, which is the same as the difference between any two corresponding terms. Why?
Solution:
Let the same common difference of two AP’s is d. Given that, the first term of first AP and second AP are 2 and 7 respectively, then the AP’s are
2, 2 + d, 2 + 2d, 2 + 3d ….
and 7, 7 + d, 7 + 2d, 7 + 3d ….
Now, 10^th terms of first and second AP’s are 2 + 9d and 7 + 9d, respectively
[∵ a₁₀ = a + (10 – 1 )d] So, their difference is 7 + 9d – (2 + 9d) = 5 Also, 21st terms of first and second AP’s are 2 + 20d and 7 + 20d, respectively
[∵ a₂₁ = a + (21 – 1 )d] So, their difference is 7 + 20d – (2 + 20d) = 5
Also, if the a_n and b_n are n^th terms of first and second AP.
Then, b_n – a_n = [7 + (n – 1 )d)] – [2 + (n – 1)d] = 5
Hence, the difference between any two corresponding terms of such AP’s is the same as the difference between their first terms.

Question 5.
Is 0 a term of the AP: 31, 28, 25, …? Justify your answer.
Solution:
Let 0 be the nth term of given AP i.e., a_n = 0
Given that, first term a = 31,
common difference, d = 28 – 31 = – 3
The n^thterms of an AP, is
a_n = a + (n – 1 )d
⇒ 0 = 31 + (n – 1)(-3)
⇒ 3(n – 1) = 31
⇒ n – 1 = 313
∴ n = 313+1=343=1113
Since, n should be positive integer. So, 0 is not a term of the given AP.

Question 6.
The taxi fare after each km, when the fare is Rs 15 for the first km and Rs 8 for each additional km, does not form an AP as the total fare (in Rs) after each km is 15, 8, 8, 8,… Is the statement true? Give reasons.
Solution:
Because the total fare (in Rs) after each km is
15, (15 + 8), (15 + 2 × 8), (15 + 3 × 8),…
= 15, 23, 31, 39,…
Let t₁ = 15, t₂ = 23, t₃ = 31 and t₄ = 39
Now, t₂ – t₁ = 23 – 15 = 8
t₃ – t₂ = 31 – 23 = 8
t₄ – t₃ = 39 – 31 = 8
Since, all the successive terms of the given list have same difference i.e., common difference is 8. Hence, the total fare after each km form an AP.

Question 7.
In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers.
(i) The fee charged from a student every month by a school for the whole session, when the monthly fee is Rs 400.
(ii) The fee charged every month by a school from Classes I to XII, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the next higher class.
(iii) The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10% per annum.
(iv) The number of bacteria in a certain food item after each second, when they double in every second.
Solution:
(i)The fee charged from a student every month by a school for the whole session is 400, 400, 400, 400,…. which forms an AP, with common difference (d) = 400 – 400 = 0

(ii) The fee charged every month by a school from classes I to XII is
250, (250 + 50), (250 + 2 × 50), (250 + 3 × 50),…
i.e., 250, 300, 350, 400,…
which forms an AP, with common difference (d) = 300 – 250 = 50

(iv) Let the number of bacteria in a certain food = x
Since, they double in every second .-. x, 2x, 2(2x), 2(2 . 2 . x),…
i.e., x, 2x, 4x, 8x,…
Now, let t₁ = x, t₂ = 2x, t₃ = 4x and t₄ = 8x
t₂ – t₁ = 2x – x = x
t₃ – t₂ = 4x – 2x = 2x
t₄ – t₃ = 8x – 4x = 4x
Since, the difference between each successsive term is not same. So, the list does form an AP.

Question 8.
Justify whether it is true to say that the following are the n^th terms of an AP.
(i) 2n – 3
(ii) 3n² + 5
(iii) 1 + n + n²
Solution:
Yes. Here a_n = 2n – 3
Put n = 1, a₁ = 2(1) – 3 = -1
Put n = 2, a₂ = 2(2) – 3 = 1
Put n = 3, a₃ = 2(3) – 3 = 3
Put n = 4, a₄ = 2(4) – 3 = 5
List of number becomes -1, 1, 3, 5….
Here, a₂ – a₁ = 1 – (-1) = 1 + 1 = 2
a₃ – a₂ = 3 – 1 = 2
a₄ – a₃ = 5 – 3 = 2
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = …………
Hence, 2n – 3 is the n^th term of an AP.

(ii) No. Here a_n = 3n² + 5
Put n = 1, a₁ = 3(1)² + 5 = 8
Put n = 2, a₂ = 3(2)² + 5 = 3(4) + 5 = 17
Put n = 3, a₁ = 3(3)² + 5 = 3(9) + 5 = 27 + 5 = 32
So, the list of number becomes 8, 17, 32,….
Here, a₂ – a₁ = 17 – 8 = 9
a₃ – a₂ = 32 -17 = 15
∴ a₂ – a₁ ≠ a₃ – a₂
Since, the difference of the successive term is not same,. So, it does not form an AP.

(iii) No. Here an = 1 + n + n²
Put n = 1, a₁ = 1 + 1 + (1)² = 3
Put n = 2, a₂ = 1 + 2 + (2)² = 1 + 2 + 4 = 7
Put n = 3, a₃ = 1 + 3 + (3)² = 1+ 3 + 9 = 13
So, the list of number becomes 3, 7, 13,…
Here, a₂ – a₁ = 7 – 3 = 4
a₃ – a₂ = 13 – 7 = 6
∴ a₂ – a₁ ≠ a₃ – a₂
Since, the difference of the successive term is not same. So it does not form an AP.

NCERT Exemplar Class 10 Maths Chapter 5 Exercise – 5.3

(iv) Here, a₁ = a + 6, a₂ = (a + 1) + b,
a₃ = (a+ 1) + (b +1)
a₂ – a₁ = (a + 1) + b – (a + b) = a + 1 + b – a – b = 1
a₃ – a₂ = (a + 1) + (b + 1) — [(a + 1) + b] = a + 1 + b + 1 – a – 1 – b = l
∵ a₂ – a₁ = a₃ – a₂ = 1
Therefore, the each successive term of the given list has same difference.
So, it forms an AP.
The next three terms are,
a₄ =a₁ + 3d = a + b + 3(1) = (a + 2) + (b + 1)
a₅ = a₁ + 4d = a + b + 4(1) = (a + 2) + (b + 2)
a₆ = a₁ + 5d = a + b + 5(1) = (a + 3) + (b + 2)

(v) Here, a₁ = a, a₂ = 2a + 1, a₃ = 3a + 2 and a₄ = 4a + 3
a₂ – a₁ = 2a + 1 – a = a + 1
a₃ – a₂ = 3a + 2 – 2a – 1 = a + 1
a₄ – a₃ = 4a + 3 – 3a – 2 = a + 1
∵ a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = a + 1
Therefore, the each successive term of the given list has same difference.
So, it forms an AP.
The next three terms are,
a₅ = a + 4d = a + 4 (a + 1) = 5a + 4
a₆ = a + 5d = a + 5(a + 1) = 6a + 5
a₇ = a + 6 d = a + 6 (a + 1) = 7a + 6

Question 4.
Find a, b and c such that the following numbers are in AP: a, 7, b, 23, c.
Solution:
Since a, 7, b, 23, c are in AP.
∴ 7 – a = b – 7 = 23 – b = c – 23
Taking second and third terms, we get
b – 7 = 23 – b
⇒ 2b = 30
∴ b = 15
Taking first and second terms, we get
7 – a = b – 7
⇒ 7 – 8 = 15 – 7 [∵ b = 15] ⇒ 7 – a = 8
∴ a = -1
Taking third and fourth terms, we get
23 – b = c – 23
⇒ 23 – 15 = c – 23 [vh 15] ⇒ 8 = c – 23
⇒ 8 + 23 = c ⇒ c = 31
Hence, a = -1, & = 15, c = 31

Question 5.
Determine the AP whose fifth term is 19 and the difference of the eighth term from the thirteenth term is 20.
Solution:
Let the first term of an AP be a and common difference d.
Given, a₅ = 19 and a₁₃ – a₈ = 20 [given] ∴ a₅ = a + (5 – 1)d = 19 and
[a + (13 – 1 )d] – [a + (8 – 1 )d] = 20
[∵ a_n = a + (n- 1 )d] ⇒ a + 4d = 19 …………. (i)
and a + 12d – a – 7d = 20 ⇒ 5d = 20
∴ d = 4
On putting d = 4 in Eq. (i), we get
a + 4(4) = 19
8 + 16 = 19
8 = 19 – 16 = 3
So, required AP is a, a + d, a + 2d, a + 3d …. i.e.,
3, 3 + 4, 3 + 2(4), 3 + 3(4),…. i.e., 3, 7, 11, 15, …

Question 6.
The 26^th, 11^th and the last term of an AP are 0, 3 and −15 respectively. Find the common
difference and the number of terms.
Solution:
Let the first term, common difference and number of terms of an AP are a, d and n, respectively.
We know that, if last term of an AP is known, then
l = 8 + (11-1 )d …………. (i)
and n^th term of an AP is
T_n = a + (n – 1)d …………. (ii)
Given that, 26^th term of an AP = 0
⇒ T₂₆ = a + (26 – 1 )d = 0 [from Eq.(i)] ⇒ 8 + 25d = 0 …………. (iii)
11^th term of an AP = 3
⇒ T₁₁ = s + (11 – 1)d = 3 [from Eq.(ii)] ⇒ 8 + 10d = 3 ……………… (iv)
and last term of an AP = -1/ 5
⇒ l = a + (n – 1 )d [from Eq (i)] ⇒ -1/5 = a + (n – 1 )d ………… (v)
Now, subtracting Eq(iv) from Eq(iii),
15 d = – 3
⇒ d = −15
Put the value of d in Eq.(iii), we get
a + 25(−15) = 0
⇒ a – 5 = 0 ⇒ 8 = 5
Now put the value of a, d in Eq. (v), we get
-1/5 = 5 + (n – 1)(-1/5)
⇒ -1 = 25 – (n – 1)
⇒ -1 = 25 – n + 1
⇒ n = 25 + 2 = 27
Hence, the common difference and number of terms are -1/ 5 and 27, respectively.

Question 7.
The sum of the 5^th and the 7^th terms of an AP is 52 and the 10th term is 46. Find the AP.
Solution:
Let the first term and common difference of AP are a and d, respectively.
According to the question,
a₅ + a₇ = 52 and a₁₀ = 46
⇒ a + (5 – l)d + a + (7 – 1)d = 52
[∵ a_n = a + (n- 1 )d] and a + (10 – 1 )d = 46
⇒ a + 4d + a + 6d = 52
and a + 9d = 46
⇒ 2a + 10d = 52
and a + 9d = 46
⇒ a + 5d = 26 ………….. (i)
a + 9d = 46 ……………. (ii)
On subtracting Eq. (i) from Eq. (ii), we get
4d = 20 ⇒ d = 5 From Eq. (i) a = 26 – 5(5) = 1
So, required AP is a, a + d, a + 2d, a + 3d …. i.e., 1, 1 + 5, 1 + 2(5), 1 + 3(5)… i.e., 1, 6,11,16,….

Question 8.
Find the 20^th term of the AP whose 7^th term is 24 less than the 11th term, first term being 12.
Solution:
Let the first term, common difference and number of terms of an AP are a, d and n, respectively.
Given that, first term (a) = 12 Now by given condition,
7^th term (T₇) = 11th term (T₁₁) – 24
[∵ n^th term of an AP, T_n = a + (n – 1 )d] ⇒ a + (7 – 1)d = a + (11 – l)d – 24
⇒ a + 6d = a + 10d – 24
⇒ 24 = 4d ⇒ d = 6
∴ 20^th term of AP, T₂₀ = a + (20 – 1)d
= 12 + 19 × 6 = 126
Hence, the required 20^th term of an AP is 126.

Question 9.
If the 9^th term of an AP is zero, prove that its 29^th term is twice its 19^th term.
Solution:
Let the first term, common difference and number of terms of an AP are a, d and n respectively.
Given that, 9^th term of an AP, i.e., T₉ = 0
[∵ n^th term of an AP, T_n = a + (n – 1 )d] ⇒ a + (9 – 1)d = 0
⇒ a + 8d = 0 ⇒ a = -8 d ………….. (i)
Now, its 19^th term, T₁₉ = a + (19 – 1)d
= -8d + 18d [From Eq.(i)] = 10d …………… (ii)
and its 29^th term, T₂₉ = a + (29 – 1)d
= -8d + 28d = 20d = 2 × (10d)
[from Eq.(i)] ⇒ T₂₉ = 2 × T₁₉
Hence, its 29th term is twice its 19th term.

Question 10.
Find whether 55 is a term of the AP: 7, 10, 13,— or not. If yes, find which term it is.
Solution:
Yes. Let the first term, common difference and the number of terms of an AP are a, d and n respectively.
Let the n^th term of an AP be 55. i.e., T_n = 55
We know that, the n^th term of an AP,
T_n = a + (n – 1 )d …………….. (i)
Given that, first term (a) = 7
and common difference (d) = 10 – 7 = 3
From Eq. (i), 55 = 7 + (n – 1) × 3
⇒ 55 = 7 + 3M – 3 = 4 + 3n
⇒ 3n = 51
∴ n = 17
Since, n is a positive integer. So 55 is a term of the AP.
Since, n = 17
Therefore, 17^th term of an AP is 55.

Question 11.
Determine k so that k²+ 4k + 8, 2k² + 3k + 6, 3k² + 4k + 4 are three consecutive terms of an AP.
Solution:
Since, k² + 4k + 8, 2k² + 3k + 6 and 3k² + 4k+ 4 are consecutive terms of an AP.
∴ 2k²+ 3k + 6- (k² + 4k + 8)
= 3k² + 4k + 4 – (2k² + 3k + 6) = Common difference
⇒ 2k² + 3k + 6 – k² – 4k – 8 = 3k² + 4k + 4 – 2k² – 3k – 6
⇒ k² – k – 2 = k² + k – 2
⇒ -k = k ⇒ 2k = 0 ⇒ k = 0

Question 12.
Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.
Solution:
Let the three parts of the number 207 are (a – d), a and (a + d), which are in AP.
Now, by given condition,
⇒ Sum of these parts = 207
⇒ a – d + a + a + d = 207
⇒ 3a = 207
a = 69
Given that, product of the two smaller parts = 4623
⇒ a(a -d) = 4623
⇒ 69 . (69 – d) = 4623
⇒ 69 – d = 67
⇒ d = 69 – 67 = 2
So, first part = a – d = 69 – 2 = 67,
Second part = a = 69
and third part = n + d = 69 + 2 = 71
Hence, required three parts are 67, 69, 71.

Question 13.
The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
Solution:
Given that, the angles of a triangle are in AP.
Let A, B and C are angles of a ∆ABC
B = A+C2
⇒ 2B = A + C …(i)
We know that, sum of all interior angles of a ∆ABC is 180° ’
A + B + C = 180°
⇒ 2B + B = 180° [from Eq(i)] ⇒ 3B = 180°
⇒ B = 60°
Let the greatest and least angles are A and C respectively
A = 2C [by condition] …………. (ii)
Now, put the values of B and A in Eq(i), we get
2 × 60° = 2C + C
⇒ 120° = 3 C
⇒ C = 40°
Put the value of C in Eq(ii), we get
A = 2 × 40°
⇒ A = 80°
Hence, the required angles of a triangle are 80°, 60° and 40°.

Question 15.
If sum of the 3^rd and the 8^th terms of an AP is 7 and the sum of the 7^th and the 14^th terms is -3, find the 10^th term.
Solution:
Let the first term and common difference of an AP are a and d, respectively.
According to the questiuon,
a₃ + a₈ = 7 and a₁₇ + a₁₄ = -3
⇒ a + (3 – 1)d + a + (8 – 1)d = 7
[∵ a_n = a + (n- 1 )d] and a + (7 – 1 )d + a + (14 – 1 )d = -3
⇒ a + 2d + a + 7d = 7
and a + 6d + a + 13d = -3
⇒ 2A + 9d = 7 ………….. (i)
and 2a + 19d = -3 …(ii)
On subtracting Eq(i) from Eq(ii), we get
10d = -10 ⇒ d = -1
2a + 9(-1) = 7 [from Eq(i)] ⇒ 2a – 9 = 7
⇒ 2a = 16
⇒ a = 8
∴ a₁₀ = a + (10 – 1)d
= 8 + 9(-1) = 8 – 9 = -1

Question 16.
Find the 12^th term from the end of the AP: -2, -4,-6, …,-100.
Solution:
Given AP : -2, -4, -6,…., -100
Here, first term (a) = -2,
common difference (d) = -4 – (-2) = -2
and the last term (l) = -100
We know that, the n^th term of an AP from the
end a_n = l – (n – 1 )d, where l is the last term
and d is the common difference.
∴ 12^th term from the end, i.e.,
a₁₂ = -100 – (12 – 1)(-2)
= -100 + (11)(2) = -100 + 22 = -78
Hence, the 12th term from the end is -78

Question 17.
Which term of the AP: 53, 48, 43,… is the first negative term?
Solution:
Given AP is 53, 48, 43,…
whose first term (a) = 53 and
common difference (d) = 48 – 53 = -5
Let n^th term of the AP be the first negative term.
i.e., T_n < 0
[∵ n^th term of an AP, T_n = a + (n – 1)d] ⇒ [a + (n – 1 )d] < 0
⇒ 53 + (n – 1)(-5) < 0
⇒ 53 – 5n + 5 < 0
⇒ 58 – 5n < 0 ⇒ 5n > 58
⇒ n > 11.6 ⇒ n = 12
i.e., 12^th term is the first negative term of the given AP
∴ T₁₂ = a + (12 – 1)d = 53 + 11 (-5)
= 53 – 55 = – 2 < 0

Question 18.
How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 31
Solution:
Here, the first number is 11, which divided by 4 leave a remainder 3 between 10 and 300
Last term before 300 is 299, which divided by 4 leave remainder 3.
∴ 11, 15, 19, 23, 299
Here, first term (a) = 11,
common difference (d) = 15 – 11 = 4
[∵ n^th term, a_n = a + (n – 1)d] ⇒ 299 = 11 + (n – 1) 4
⇒ 4(n – 1) = 288
⇒ (n – 1) = 72
∴ n = 73

Question 20.
The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Solution:
Let the first term, common difference and the number of terms of an AP are a, d and n respectively.
Given that, first term (a) = -5 and
last term (l) = 45
Sum of the terms of the AP = 120 ⇒ S_n = 120
We know that, if last term of an AP is known, then sum of n terms of an AP is,
S_n = n2 (a + l)
⇒ 120 = n2(-5 + 45) ⇒ 120 × 2 = 40 × n
⇒ n = 3 × 2 ⇒ n = 6
∴ Number of terms of an AP is known, then the n^th term of an AP is,
l = a + (n – 1)d ⇒ 45 = -5 + (6 – 1)d
⇒ 50 = 5d ⇒ d = 10
So, the common difference is 10.
Hence, number of terms and the common difference of an AP are 6 and 10 respectively.

Question 22.
Which term of the AP: -2, -7, -12,… will be -77? Find the sum of this AP upto the term -77.
Solution:
Given, AP : -2, -7, -12, ….
Let the n^th term of an AP is -77
Then, first term (a) = -2 and
common difference (d) = -7 – (-2) = -7 + 2 = -5
∵ n^th term of an AP, T_n = a + (n – 1)d
⇒ -77 = -2 + (n – 1)(-5)
⇒ -75 = -(n – 1) × 5
⇒ (n – 1) = 15 ⇒ n = 16
So, the 16th term of the given AP will be -77
Now, the sum of n terms of an AP is
S_n = n2 [2a + (n – 1)d] So, sum of 16 terms i.e., upto the term -77
S₁₆ = 162 [2 × (-2) + (n – 1)(-5)l
= 8[-4 + (16 – 1)(-5)] = 8(-4 – 75)
= 8 × (-79)= -632
Hence, the sum of this AP upto the term -77 is -632.

Question 23.
If an = 3 – 4n, show that a₁, a₂, a₃,… form an AP. Also find S₂₀
Solution:
Given that, n^th term of the series is
a_n = 3 – 4n …(i)
Put n = 1, a₁ = 3 – 4(1) = 3 – 4 = -1
Put n = 2, a₂ = 3 – 4(2) = 3 – 8 = -5
Put n = 3, a₃ = 3 – 4(3) = 3 – 12 = -9
Put n = 4, a₄ = 3 – 4(4) = 3 – 16 = -13
So, the series becomes -1, -5, -9, -13,….
We see that,
a₂ – a₁ = -5 – (-1) = -5 + 1 = -4,
a₃ – a₂ = -9- (-5) = -9 + 5 = -4,
a₄ – a₃ = -13 – (-9) = -13 + 9 =-4
i.e., a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = … = —4
Since, the each successive term of the series has the same difference. So, it forms an AP. We know that, sum of n terms of an AP,
S_n = n2 [2a + (n – 1)d] ∴ Sum of 20 terms of the AP,
S₂₀ = 202 [2(-1) + (20 – 1)(-4)] = 10 [-2 + (19)(-4)] = 10(-2 – 76)
= 10 × (-78) = -780
Hence, the required sum of 20 terms i.e., S20 is -780

Question 24.
In an AP, if S_n = n (4n + 1), find the AP.
Solution:
We know that, the n^th term of an AP is
a_n = S_n – S_{n – 1}
a_n = n(4n + 1) – (n – 1) {4(n -1) + 1}
[∵ S_n = n (4n + 1)] ⇒ a_n = 4n² + n- (n – 1)(4 n – 3)
= 4n² + n – 4n² + 3n + 4n – 3 = 8n – 3
Put n = 1, a₁ = 8(1) -3 = 5
Put n = 2, a₂ = 8(2) – 3 = 16 – 3 = 13
Put n = 3, a₃ = 8(3) – 3 = 24 – 3 = 21
Hence, the required AP is 5, 13, 21,….

Question 25.
In an AP, if S_n = 3n² + 5n and a_k = 164, find the value of k.
Solution:
∵(n^th term of an AP,
a_n = S_n – S_{n – 1}
= 3n² + 5n – 3(n – 1)² – 5(w – 1)
[∵ S_n = 3n² + 5n (given)] = 3n² + 511 – 3n² – 3 + 6n – 5n + 5
⇒ a_n = 6n + 2 …………… (i)
or a_k = 6k + 2 = 164 [∵ a_k = 164 (given)] ⇒ 6k = 164 – 2 = 162
∴ k = 27

Question 30.
Find the sum of last ten terms of the AP: 8,10, 12,……., 126.
Solution:
For finding the sum of last ten terms, we write the given AP in reverse order.
i.e., 126, 124, 122,…., 12, 10, 8
Here, first term (a) = 126,
common difference (d) = 124 – 126 = – 2
∴ S₁₀ = 102 [2a + (10 – 1)d] [∵ S_n = n2 [2a + (n – 1)d]] = 5{2(126) + 9(—2)} = 5(252 – 18)
= 5 × 234 = 1170

Question 32.
How many terms of the AP: -15, -13, -11,— are needed to make the sum -55? Explain the reason for double answer.
Solution:
Let n number of terms are needed to make the sum -55
Here, first term (a) = -15,
common difference (d) = -13 + 15 = 2
∵ Sum of n terms of an AP,
S_n = n2 [2a + (n – 1)d] ⇒ -55 = n2 [2(-15) + (n – 1)2] [∵ S_n = -55 (given)] ⇒ -55 = -15M + n(n – 1)
⇒ n² – 16n + 55 = 0
⇒ n² – 11n – 5n + 55 = 0
[by factorisation method] ⇒ n(n – 11) – 5(n – 11) = 0
⇒ (n – 11)(n – 5) = 0
∴ n = 5, 11
Hence, either 5 or 11 terms are needed to make the sum -55 when n = 5,
AP will be -15, -13, -11, -9, -7,
So, resulting sum will be -55 because all terms are negative.
When n = 11,
AP will be -15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5
So resulting sum will be -55 because the sum of terms 6th to 11th is zero.

Question 35.
Yasmeen saves ₹ 32 during the first month, ₹ 36 in the second month and ₹ 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?
Solution:
Given that,
Yasmeen, during the first month, saves = ₹ 32
During the second month, saves = ₹ 36
During the third month, saves = ₹ 40
Let Yasmeen saves Rs 2000 during the n months.
Here, we have arithmetic progression 32, 36, 40,…
First term (a) = 32,
common difference (d) = 36 – 32 = 4
and she saves total money, i.e., S_n = ₹ 2000
We know that, sum of first n terms of an AP is,
S_n = n2[2a + (n – 1)d] ⇒ 2000 = n2 [2 × 32 + (n -1) × 4] ⇒ 2000 = n(32 + 2n – 2)
⇒ 2000 = n(30 + 2n)
⇒ 1000 = n (15 + n)
⇒ 1000 = 15n + n²
⇒ n² + 15n – 1000 = 0
⇒ n² + 40n – 25n – 1000 = 0
⇒ n(n + 40) – 25(n + 40) = 0
⇒ (n + 40)(n – 25)= 0
∴ n = 25 [ ∵ n ≠ – 40] [since, months cannot be negative] Hence, in 25 months she will save ₹ 2000.

NCERT Exemplar Class 10 Maths Chapter 5 Exercise – 5.4

Question 2.
Find the
(i) Sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
(ii) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
(iii) sum of those integers from 1 to 500 which are multiples of 2 or 5.
Solution:
(i) Since, multiples of 2 as well as of 5
= LCM of (2, 5) = 10
∴ Multiples of 2 as well as of 5 between 1
and 500 is 10, 20, 30,…, 490
which form an AP with first term (a) = 10 and
common difference (d) = 20 – 10 = 10
n^th term a_n = Last term (l) = 490
∴ Sum of n terms between 1 and 500,
S_n = n2 [a + l] …………. (i)
[∵ a_n = a + (n – 1)d = l] ⇒ 10 + (n – 1) 10 = 490
⇒ (n -1)10 = 480
⇒ n — 1 = 48 ⇒ n = 49
From Eq (i), S₄₉ = 492 (10 + 490)
= 492 × 500 = 49 × 250 = 12250

(ii) Same as part (i),
But multiples of 2 as well as 5 from 1 to 500 is 10, 20, 30,…, 500.
∴ a = 10, d = 10, a_n = l = 500
∵ a_n = a + (n – 1)d = 1
⇒ 500 = 10 + (n- 1)10
⇒ 490 = (n – 1) 10
⇒ n – 1 = 49
⇒ n = 50
∵ S_n = n2 (a + l)
⇒ S₅₀ = 502 (10 + 500) = 502 × 510
= 50 × 255 = 12750

Question 3.
The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15^th term.
Solution:
Let a and d the first term and common difference of an AP, respectively
Now, by given condition, a₈ = 12 a₂
⇒ a + 7d = 12 (a + d) [∵ a_n = a + (n – 1 )d] ⇒ 2a + 14d = a + d
⇒ a + 13d = 0 …………. (i)
and a₁₁ = 13 a₄ + 1 [given] ⇒ a + 10d = 13 [a – 3d] + 1
⇒ 3 a + 30d = a + 3d + 3
⇒ 2a + 27d = 3
From Eqs. (i) and (ii)
2(-13d) + 27d = 3
⇒ -26d + 27d = 3
⇒ d = 3
From Eq. (i),
a + 13(3) = 0
⇒ a = -39
∴ a₁₅ = a + 14d = -39 + 14(3) = -39 + 42 = 3

Question 4.
An AP consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the last three is 429. Find the AP.
Solution:
Since, total number of terms (n) = 37 [odd] ∴ Middle term = (37+12)th term = 19th term
So, the three middle most terms = 18^th, 19^th and 20^th,
By given condition.
Sum of the three middle most terms = 225
a₁₈ + a₁₉ + a₂₀ = 225
⇒ (a + 17d) + (a + 18d) + (a + 19 d) = 225
[∵ a_n = a + (n – 1 )d] ⇒ 3a + 54d = 225
⇒ a + 18d = 75 …………. (i)
and sum of the last three terms = 429
⇒ a₃₅ + a₃₆ + a₃₇ = 429
⇒ (a + 34d) + (a + 35d) + (a + 36 d) = 429
⇒ 3a + 105d = 429
⇒ a + 35d = 143 ………….. (ii)
On subtracting Eq.(i) from Eq. (ii), we get
17d = 68
⇒ d = 4
From Eq. (i), a + 18(4) = 75
⇒ a = 75 – 72
⇒ a = 3
∴ Required AP is a, a + d, a + 2d, a + 3d,….
i.e., 3, 3 + 4, 3 + 2(4), 3 + 3(4),…
i.e., 3, 7, 3 + 8, 3 + 12,….
i.e., 3, 7, 11, 15,….