Question

The sum of first n terms of an AP is given by 2
n S = 2n + 3n . Find the sixteenth term of
the AP.

Answer 1

A correction in your question :The sum of first n terms of an AP is given by S = 2n² + 3n . Find the sixteenth term of the AP.


Given,
Sum of first n terms = 2n² + 3n

Now,

Sum of first 1 terms = First term = 2(1²)+3(1) = 5 .
Sum of first two terms = 2(2²)+3(2)=14 .

Second term = 14 - 5 = 9 .

Now,
Common difference = 9 - 5 = 4 .

We have, a = 5 , d = 4 .

Sixteen th term of the A. P = 5 + ( 16 - 1 ) 4 = 5 + 15(4) = 5 + 60 = 65 .

Hope this helps!

Answer 2

Heya !!

=================================

Given :- Sum of first n terms of an AP = 2n²+3n

S1 = 2(1)² + 3(1)
=> 5

S2 = 2(2)² + 3(2)
=> 8 + 6
=> 14

S13 = 2(3)² + 3(3)
=> 18 + 9
=> 27

Now, a1 = 5

a2 = S2 – S2
=> 14 – 5
=> 9

a3 = S3 – S2
=> 27 – 14
=> 13

We have, a = 5,
d = 4 and
n = 16

a16 = a + 15d
=> 5 + 15(4)
=> 5 + 60
=> 65

Hope my ans.'s satisfactory.☺