ICSE Solutions for Chapter 10 Arithmetic Progression Class 10 Mathematics
Solution 1: The given A.P. is 8, 3, -2......
Here, a = 8, d = 3 – 8 = - 5 and n = 22
∴ S = n/2[2a + (n - 1)d]
= 22/2 [2×8+ (22 - 1) × (-5)]
= 11[16 + 21×(-5)]
= 11[16 – 105]
= 11 × (-89)
= -979
Question 2: How many two-digit numbers are divisible by 3?
Solution 2: The two-digit numbers divisible by 3 are as follows:
12, 15, 18, 21, ...., 99
Clearly, this forms an A.P. with first term, a = 12
and common difference, d = 3
Last term = nth term = 99
The general term of an A.P. is given by
tn = a + (n - 1)d
⇒ 99 = 12 + (n - 1)(3)
⇒ 99 = 12 + 3n - 3
⇒ 90 = 3n
⇒ n = 30
Thus, 30 two-digit numbers are divisible by 3.
Question 3: If the pth term of an A.P.is (2p+ 3), find the A.P.
Solution 3: pth term of an A.P. = 2p + 3
⇒ tp = 2p + 3
Putting t = 1, 2, 3, .... we get
t1 = 2×1 + 3 = 2 + 3 = 5
t2 = 2×2 + 3 = 4 + 3 = 7
t3 = 2×3 + 3 = 6 + 3 = 9 and so on.
Thus, the A.P. is 5, 7, 9, …
Question 4: Find the 24th term of the sequence: 12, 10.8,6.....
Solution 4:
The given sequence is 12, 10, 8, 6,....
Now,
10 - 12 = -2
8 – 10 = -2
6 – 8 = -2, etc.
Hence, the given sequence is an A.P. with first term a = 12
and common difference d = -2.
The general term of an A.P. is given by
tn = a + (n - 1)d
⇒ t24 = 12 + (24 – 1)(-2) = 12 + 23×(-2) = 12 - 46 = -34
So, the 24th term is - 34.
Question 5: If the 3rd and the 9th terms of an arithmetic progression are 4 and -8 respectively, Which term of it is zero?
Solution 5: For an A.P.,
t3 = 4
⇒ a + 2d = 4 ...(i)
t9 = -8
⇒ a + 8d = - 8 ...(ii)
Subtracting (i) from (ii), we get
6d = -12
⇒ d = -2
Substituting d = -2 in (i), we get
a + 2(-2) = 4
⇒ a – 4 = 4
⇒ a = 8
⇒ General term = tn = 8 + (n - 1)(-2)
Let pth term of this A.P. be 0.
⇒ 8 + (p - 1) × (-2) = 0
⇒ 8 – 2p + 2 = 0
⇒ 10 - 2p = 0
⇒ 2p = 10
⇒ p = 5
Thus, 5th term of this A.P. is 0.
Question 6: The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms
Solution 6: Sum of first 7 terms of an A.P = 49
⇒ 7/2[2a + 6d] = 49
⇒ 7/2 ×2[a + 3d] = 49
⇒ 7[a + 3d] = 49
⇒ a + 3d = 7 ...(i)
Sum of first 17 terms of A.P. = 289
⇒ 17/2[2a + 16d] = 289
⇒ 17/2 × 2 [a + 8d] = 289
⇒ 17[a + 8d] = 289
⇒ a + 8d = 17 ... (ii)
Subtracting (i) from (ii), we get
5d = 10 ⇒ d = 2
Substituting d = 2 in (i), we get
a + 3×2 = 7
⇒ a + 6 = 7 ⇒ a = -1
∴ Sum of first n terms = n/2 [2×1 + (n – 1)2]
= n/2[2 + 2n - 2]
= n/2 × 2n
= n2
Question 7: For what value of n, the nth term of A.P 63, 65, 67,.....and nth term of A.P. 3, 10, 17,... are equal to each other?
Solution 7: For an A.P. 63, 65, 67, ....., we have a = 63 and d = 65 – 63 = 2
nth term = tn = 63 + (n - 1)×2
For an A.P. 3, 10, 17, ......, we have a' = 3 and d' = 10 – 3 = 7
nth term = t'n = 3 + (n - 1) x 7
The two A.P.s will have equal nth terms is
tn = t’n
⇒ 63 + (n - 1)×2 = 3 + (n - 1)×7
⇒ 63 + 2n – 2 = 3 + 7n = 7
⇒ 61 + 2n = 7n - 4
⇒ 5n = 65
⇒ n = 13
Question 8: Is 402 a term of the sequence:
8, 13, 18, 23 .......?
Solution 8:
The given sequence is 8, 13, 18, 23,...
Now,
13 – 8 = 5
18 – 13 = 5
23 – 18 = 5, etc
Hence, the given sequence is an A.P. with first term a = 8
and common difference d = 5.
The general term of an A.P. is given by
tn = a + (n – 1)d
⇒ 402 = 8 + (n - 1)(5)
⇒ 394 = 5n - 5
⇒ 399 = 5n
⇒ n = 399/5
The number of terms cannot be a fraction.
So clearly, 402 is not a term of the given sequence.
Question 9: If numbers n - 2, 4n - 1 and 5n + 2 are in A.P. find the value of n and its next two terms.
Solution 9: Since (n - 2), (4n - 1) and (5n + 2) are in A.P., we have
(4n - 1) - (n - 2) = (5n + 2) - (4 - 1)
⇒ 4n - 1 – n + 2 = 5n + 2 – 4n + 1
⇒ 3n + 1 = n + 3
⇒ 2n = 2
⇒ n = 1
So, the given numbers are = -1, 3, 7
⇒ a = -1 and d = 3 - (-1) = 4
Hence, the next two terms are (7 + 4) and (7 + 2×4)
i.e. 11 and 15.
Question 10: In an A.P. the first term is 25, nth term is -17 and the sum of n terms is 132. Find n and the common difference.
Solution 10: First term a = 25
nth term = -17
⇒ Last term I = -17
Sum of n terms = 132
⇒ n/2 [a + l] = 132
⇒ n(25 - 17) = 264
⇒ n×8 = 264
⇒ n = 33
Now, I = -17
⇒ a + (n - 1)d = -17
⇒ 25 + 32d = -17
⇒ 32d = -42
⇒ d = - 42/32
⇒ d = -21/16
Sum of n terms = 132
⇒ n/2 [a + l] = 132
⇒ n(25 - 17) = 264
⇒ n×8 = 264
⇒ n = 33
Now, I = -17
⇒ a + (n - 1)d = -17
⇒ 25 + 32d = -17
⇒ 32d = -42
⇒ d = - 42/32
⇒ d = -21/16