Question

The sum of first n terms of an ap is (3n^2+6n) then the common difference is

Answer 1

Answer:

6

Step-by-step explanation:

It is given that

Sum (Sn) = 3n²+6n

Substituting n as 1 we get

S1= 3(1)²+6(1)

= 3 + 6 = 9 = a1 [as sum of first term is equal to the first term]

Similarly substituting n as 2 we get

S2 = 3(2)²+6(2)

=3(4)+12

=12+12

= 24

But S2 = a1 + a2

Or 24 = 9 + a2

Therefore a2 = 24-9=15

Now you know a1 and a2 both

Therefore common difference d =

a2 - a1

= 15-9

=6

(I did this question yesterday in my online exam :)