Class 10 Maths Chapter 5 Question Answers - Arithmetic Progressions

Q1: Find the sum of all multiples of 7 lying between 500 and 900.
Ans: To find: 504 + 511 + 518 + … + 896
a = 504, d = 511- 504 = 7, a_n = 896
a + (n – 1)d = a_n
∴ 504 + (n – 1)7 = 896
(n – 1)7 = 896 – 504 = 392


Q2: The sums of first n terms of three arithmetic progressions are S₁ S₂ and S₃ respectively. The first term of each A.P. is 1 and their common differences are 1, 2 and 3 respectively. Prove that S₁ + S₃ = 2S₂.
Ans:


Q3: 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.
Ans: Let hundred’s place digit = (a – d)
Let ten’s place digit = a
Let unit’s place digit = a + d
According to the Question,
a – d + a + a + d = 15
⇒ 3a = 15 ⇒ a = 5
Original number
= 100(a – d) + 10(a) + 1(a + d)
= 100a – 100d + 10a + a + d
= 111a – 99d
Reversed number
= 1(a – d) + 10a + 100(a + d)
= a – d + 10a + 100a + 100d
= 111a + 99d
Now, Original no. – Reversed no. = 594
111a – 99d – (111a + 99d) = 594
-198d = 594 ⇒ d = 594/−198 = -3
∴ The Original no. = 111a – 99d
= 111(5) – 99(-3)
= 555 + 297 = 852

Q4: 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.
Ans: Let A be first term and D be the common difference of 1st A.P.
Let a be the first term and d be the common difference of 2nd A.P.

∴ Required ratio = (14m – 6) : (8m + 23)

Q5: If S_n denotes the sum of first n terms of an A.P., prove that S₃₀ = 3[S₂₀ – S₁₀].
Ans: Let a be the first term and d be the common difference of the A.P.

= 10a + 45d …(iii)
To prove, 3(S₂₀ – S₁₀) = S₃₀
= 3(20a + 190d – 10a – 45d) …[From (i), (ii) & (iii)
= 3(10a + 145d)
= 30a + 435d = S30 …Hence Proved

Q6: If the sum of the first n terms of an A.P. is 1/2 (3n² + 7n), then find its nth term. Hence write its 20th.
Ans:

Now we know,
S₁ = 27 and a₂ = S₂ – S₁
∴ a₁ = 5 = 13 – 5 = 8
Now we have,
a₁ = 5, a₂ = 8, d = a₂ – a₁ = 8 – 5 = 3
an = a + (n – 1)d = 5 + (n – 1)3
= 5 + 3n – 3 = 3n + 2 (nth term)
∴ 20th term, a₂₀ = (3 × 20) + 2 = 62

Q7: If Sn, denotes the sum of first n terms of an A.P., prove that S₁₂ = 3(S₈ – S₄).
Ans: Let a be the first term and d be the common difference of A.P.

12a + 660 = 3(8a + 28d – 4a – 6d) … [From (i), (ii) & (iii)
12a + 660 = 3(4a + 22d)
12a + 660 = 12a + 66d …Hence proved

Q8: The sum of the first seven terms of an AP is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the AP.
Ans: S₇ = 182 …[Given]


Q9: The sum of first n terms of an AP is 3n² + 4n. Find the 25^th term of this AP. 
Ans: We have, S_n = 3n² + 4n
Put n = 25,
S₂₅ = 3(25)² + 4(25)
= 3(625) + 100
= 1875 + 100 = 1975
Put n = 24,
S₂₄ = 3(24)² + 4(24)
= 3(576) + 96
= 1728 + 96 = 1824
∴ 25^th term = S₂₅ – S₂₄
= 1975 – 1824 = 151 

Q10: The n^th term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P. 
Ans: We have, a_n = -4n + 15
Put n = 1, a₁ = -4(1) + 15 = 11
Put n=2, a₂ = -4(2) + 15 = 7
∴ d = a₂ – a₁ = 7 – 11 = -4
As S_n = n/2 (2a + (n – 1)d]
∴ S₂₀ = 20/2[2(11) + (20 – 1)(-4))… [n = 20 (Given)]
= 10 (22 – 76)
= 10 (-54) = -540

Q11: 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.
Ans: First term, a = 5, Last term, a_n = 45
Let the number of terms = n
S_n = 400


Q12: 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.
Ans:
Let 1^st term = a, Common difference = d
Given: S₇ = 49, S₁₇ = 289 


Q13: 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.
Ans:
a₁₃ = 4a₃ … [Given
⇒ a + 12d = 4(a + 2d) …[∵ a_n = a + (n – 1)d
⇒ a + 12d – 4a – 8d = 0 


Q14: If 3k-2, 4k-6 and k+2 are three consecutive terms of AP, then find the value of k.
Ans:

Q15: Find the number of terms of the AP 18, 151/2, 13,…, -491/2 and find the sum of all its terms.
Ans: Here 1^st term, a = 18



Q16: Find the value of the middle term of the following A.P.: -6, -2, 2, …, 58.
Ans: Here a = -6, d = -2 -(-6) = 4, a_n = 58
As we know, a + (n – 1) d = 58
∴ -6 + (n – 1) 4 = 58
⇒ (n – 1) 4 = 58 + 6 = 64
⇒ (n – 1) = 64/4 = 16
⇒ n = 16 + 1 = 17 (odd) 


Q17: If the seventh term of an AP is 1/9 and its ninth term is 1/7, find its 63rd term.
Ans:


Q18: The sum of the 5th and the 9th terms of an AP is 30. If its 25^th term is three times its 8^th term, find the AP.
Ans:
a₅ + a₉ = 30 … [Given
a + 4d + a + 8d = 30 …[∵ a_n = a + (n – 1)d
2a + 12d = 30 ⇒ a + 6d = 15 …[Dividing by 2
a = 15 – 6d …(i)
Now, a₅₂ = 3(a₈)
a + 24d = 3(a + 7d)
15 – 6d + 240 = 3(15 – 6d + 7d) …[From (i)
15 + 18d = 3(15 + d)
15 + 18d = 45 + 3d
18d – 3d = 45 – 15
15d = 30 ∴ d = 30/15 = 2
From (i), a = 15 – 6(2) = 15 – 12 = 3 


Q19: The 9th term of an A.P. is equal to 6 times its second term. If its 5^th term is 22, find the A.P. 
Ans:
9^th term = 6 (2^nd term)
∴ a +8d = 6 (a + d) …[As a_n = a + (n – 1)d
a + 8d = 6a + 6d
8d – 6d = 6a – a
2d = 5a
⇒ d = 5a/2 …(i)
Now, a₅ = 22
a + 4d = 22 


Q20: The 19th term of an AP is equal to three times its 6^th term. If its 9^th term is 19, find the A.P.
Ans:
Given: a₁₉ = 3(a₆)
⇒ a + 18d = 3(a + 5d)
a + 18d = 3a + 15d
18d – 15d = 3a – a