Question

The sum of first n terms of an ap is 3n2 + 6n. find the nth term and the 15th of this ap

Answer 1

QUESTION:
The sum of first n terms of an ap is 3n2 + 6n. find the nth term and the 15th of this ap?

SOLUTION:

If n = 1, then
$= 3(1 {}^{2} ) + 6(1)$
$= 3(1) + 6(1) = 9$
The first term is 9.

If n=2 ,then

$= 3(2 {}^{2} ) + 6(2)$
$= 3(4) + 12 = 24$
The second term will be

=24-9=15.

Now, Common difference will be 15-9=6.

So, fifteen term will be

$= a + 14d$
$= 9 + 14(6) = 9 + 84 = 93$

Answer 2

Given the sum of n terms= 3n^2+6n

To find nth term and 15 th term

Sn = 3n^2+6n

S1 =3(1)^2+6(1)=3+6=9. (S1=a1)..........1

S2= 3(2)^2+6(2)

3*4+12=12+12 = 24

a2 = S2-S1

24-9=15

d = a2-a1

15-9= 6 .......(2)

an = a+(n-1)d

9+(n-1)*6

9+6n-6= 6n+3.. ..(3). ....( from 1 and 2)

Now

Sub n=15 in 3

a15 = 6*15+3

90+3

a15=93. And an = 6n+3