# 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, an = 896
a + (n – 1)d = an
∴ 504 + (n – 1)7 = 896
(n – 1)7 = 896 – 504 = 392

Q2: The sums of first n terms of three arithmetic progressions are SS2 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 Sn 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
∴ a= 5 = 13 – 5 = 8
Now we have,
a1 = 5, a= 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 25th term of this AP.
Ans: We have, Sn = 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
∴ 25th term = S25 – S24
= 1975 – 1824 = 151

Q10: The nth term of an A.P. is given by (-4n + 15). Find the sum of first 20 terms of this A.P.
Ans: We have, an = -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 Sn = 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, an = 45
Let the number of terms = n
Sn = 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 1st 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) …[∵ an = 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:

To be term of an AP the difference between two
consecutive terms must be the same.
If 3k-2, 4k-6 and k+2 are terms of an AP, then
4k -6 -(3k-2) = k + 2 - (4k-6)
⇒ 4k - 6 -3k + 2 = k + 2 -4k + 6
⇒ k - 4 = 8 - 3k
⇒ 4k = 12
⇒ k = 3
Hence, the value of k is 3.

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 1st 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, an = 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 25th term is three times its 8th term, find the AP.
Ans:

a5 + a9 = 30 … [Given
a + 4d + a + 8d = 30 …[∵ an = 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 5th term is 22, find the A.P.
Ans:

9th term = 6 (2nd term)
∴ a +8d = 6 (a + d) …[As an = 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 6th term. If its 9th 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

The document Class 10 Maths Chapter 5 Question Answers - Arithmetic Progressions is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 5 Question Answers - Arithmetic Progressions

 1. How do you find the common difference in an arithmetic progression?
Ans. To find the common difference in an arithmetic progression, subtract any term from the term that follows it. The result will be the common difference between consecutive terms.
 2. What is the formula to find the nth term of an arithmetic progression?
Ans. The formula to find the nth term of an arithmetic progression is: $a_n = a_1 + (n-1)d$, where $a_n$ is the nth term, $a_1$ is the first term, $n$ is the term number, and $d$ is the common difference.
 3. How can you determine whether a given sequence is an arithmetic progression or not?
Ans. To determine if a given sequence is an arithmetic progression, calculate the differences between consecutive terms. If these differences are constant, then the sequence is an arithmetic progression.
 4. Can an arithmetic progression have negative common difference?
Ans. Yes, an arithmetic progression can have a negative common difference. This means that each term in the progression decreases by the same value as you move along the sequence.
 5. What is the sum of the first n terms of an arithmetic progression?
Ans. The sum of the first n terms of an arithmetic progression can be calculated using the formula: $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, where $S_n$ is the sum of the first n terms, $a_1$ is the first term, $n$ is the number of terms, and $d$ is the common difference.

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