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Question
The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28th term of this A.P.
Solution
Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = \[\frac{n}{2}\][2a + (n − 1)d]
It is given that sum of the first 7 terms of an A.P. is 63.
And sum of next 7 terms is 161.
∴ Sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms
= 63 + 161 = 224
Now,
S7 = \[\frac{7}{2}\][2a + (7 − 1)d]
⇒ 63 = \[\frac{7}{2}\] (2a + 6d)
⇒ 18 = 2a + 6d
⇒ 2a + 6d = 18 ....(1)
Also,
S14 = \[\frac{14}{2}\][2a + (14 − 1)d]
⇒ 224 = 7(2a + 13d)
⇒ 32 = 2a + 13d
⇒ 2a + 13d = 32 ....(2)
On subtracting (1) from (2), we get
13d − 6d = 32 − 18
⇒ 7d = 14
⇒ d = 2
⇒ 2a = 18 − 6d [From (1)]
⇒ 2a = 18 − 6 × 2
⇒ 2a = 18 − 12
⇒ 2a = 6
⇒ a = 3
Also, nth term = an = a + (n − 1)d
⇒ a28 = 3 + (28 − 1)2
= 3 + 27 × 2
= 57
Thus, 28th term of this A.P. is 57.
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