Question

The sum of the first 'n' terms of an AP is 3n2 +5n .write AP hence find its 16 term

Answer 1

Hey mate..
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Given,

The sum of n terms of an A.P. = 3n^2 +5n

Then sum of n-1 terms = 3(n-1)^2 +5(n-1)

We know,

a (n) = S(n) - S (n-1) .......(1)

So, nth term of the AP =


a(n) = 3n^2 +5n- 3(n-1)^2 -5(n-1) [From (1)]

= 3n^2 +5n-3n^2 - 3+6n-5n+5

= 6n+2.....(2)

Now,

Putting the values of n = 1,2,3...in the Eq.(2) we get,

a(1) = 6(1) + 2 = 8

a(2) = 6(2) + 2 = 14

a(3) = 6(3) + 2 = 20

Thus,

AP:- 8,14,20.......

Here,

First term , a(1) = 8

Common Difference, d = 14 - 8 = 6

Now,

A/Q,

a (16) = a + (16 - 1) d

= a + 15d

= 8 + 15×6

= 8 + 90

= 98 // Ans:-

Hope it helps !!