Question

The sum of first n terms of an AP is given above,then the common difference of the AP is
THE CORRECT ANSWER WILL BE MARKED AS BRAINLIEST.
Answer step by step.​

Answer 1

Given :-

• Sum of first n terms of AP is (3n² + 6n)

To find :-

• Common difference of AP

Solution :-

In order to find the common difference of the AP, we must find out the first term and second term of AP.

Sum of first term of AP is it's first term. Substitute the value of n = 1 in the given equation of sum of n terms of AP.

$\tt : \implies \:Sn = (3 {n}^{2} + 6n)$

$\tt : \implies \:S1 =(3 {(1)}^{2} + 6(1))$

$\tt : \implies \:S1 =(3 + 6)$

$\large\boxed{ \red{ \tt : \implies \:S1=9}}$

Since S1 = a1, first term of AP is 9.

~Now finding the sum of first 2 terms of AP

$\tt : \implies \:Sn = (3 {n}^{2} + 6n)$

$\tt : \implies \:S2 = (3 {(2)}^{2} + 6(2))$

$\tt : \implies \:S2 = (3 (4) + 6(2))$

$\tt : \implies \:S2 = (12 + 12)$

$\large \boxed{ \purple{\tt : \implies \:S2 = 24}}$

$\textsf{Sum of first two terms = 1st term + 2nd term}$

$\dashrightarrow \tt{S2 = a1 + a2}$

$\dashrightarrow \tt{24 = 9+ a2}$

$\dashrightarrow \tt{24 - 9 = a2}$

$\large\boxed{ \green{\dashrightarrow \tt{15 = a2}}}$

We have obtained the first term and common difference of the AP, now finding the common difference by subtracting 1st term from second term.

$\textsf{Common difference of AP = a2 - a1}$

$\textsf{Common difference of AP = 15 - 9}$

$\textsf{Common difference of AP = 6}$

$\large \underline{ \pink{ \sf\therefore Common \: difference \: of \: AP \:is\: 6}}$

Answer 2

Answer:

Sn=3n^2+6n

S1=1st term=3×(1)^2+6×1=9

S2=1st term+2nd term=3×(2)^2+6×2=24

2nd term=S2-S1=24-9=15

common diff=2 nd term-1st term=15-9=6