Question

If the 10^th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its n^th term.

Answer 1

Let a be the first term and d be the common difference.

We know that, sum of first n terms = S_n = $\frac{n}{2}$[2a + (n − 1)d]
and n^th term = a_n = 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,
a_n = a + (n − 1)d
= 3 + (n − 1)2
= 3 + 2n − 2
= 1 + 2n

Thus, n^th term of this A.P. is 1 + 2n.