Question

If the sum of first n terms of an A.P. is \[\frac{1}{2}\] (3n² + 7n), then find its n^th term. Hence write its 20^th term.

Answer 1

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 n terms of an A.P. is \[\frac{1}{2}\] (3n² + 7n).

∴ First term = a =  S₁ = \[\frac{1}{2}\][3(1)2 + 7(1)] = 5.

Sum of first two terms = S₂ =  \[\frac{1}{2}\][3(2)² + 7(2)] = 13.

∴ Second term = S₂ − S₁ = 13 − 5 = 8.

∴ Common difference = d = Second term − First term

= 8 − 5 = 3

Also, n^th term = a_n = a + (n − 1)d

⇒ a_n = 5 + (n − 1)(3)

⇒ a_n = 5 + 3n − 3

⇒ a_n = 3n + 2

Thus, n^th term of this A.P. is 3n + 2.

Now,

a₂₀ = a + (20 − 1)d

⇒ a₂₀ = 5 + 19(3)

⇒ a₂₀ = 5 + 57

⇒ a₂₀ = 62
Thus, 20^th term of this A.P is 62.