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Question
If the 10th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its nth term.
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Solution
Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = \[\frac{n}{2}\]
and nth term = an = a + (n − 1)d
Now,
S10 = \[\frac{10}{2}\][2a + (10 − 1)d]
⇒ 120 = 5(2a + 9d)
⇒ 24 = 2a + 9d
⇒ 2a + 9d = 24 ....(1)
Also,
a10 = a + (10 − 1)d
⇒ 21 = a + 9d
⇒ 2a + 18d = 42 ....(2)
Subtracting (1) from (2), we get
18d − 9d = 42 − 24
⇒ 9d = 18
⇒ d = 2
⇒ 2a = 24 − 9d [From (1)]
⇒ 2a = 24 − 9 × 2
⇒ 2a = 24 − 18
⇒ 2a = 6
⇒ a = 3
Also,
an = a + (n − 1)d
= 3 + (n − 1)2
= 3 + 2n − 2
= 1 + 2n
Thus, nth term of this A.P. is 1 + 2n.
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