Question

If the sum of first n terms of an A.P is (3n2+6n), then find the common difference of A.P. cbse 2014

Answer 1

Answer:

Step-by-step explanation:

Sn = 3n^2 + 6n

=> n/2 [ 2a + (n-1) d] = n (3n + 6)

=> [ 2a + (n-1) d] = 2 (3n +6)

=> 2a + (n-1) d = 6n + 12

=> 2a + (n-1)d = 18 + 6n - 6

=> 2a + (n-1)d = 18 + 6(n-1)

On Comparing both sides, we get

2a = 18

=> a = 9

And, d = 6

Tn = a + (n-1) d

= 9 + (n-1) 6

= 9 + 6n - 6

Answer 2

Answer:

d=6

Step-by-step explanation:

Sn=n/2+[2a+(n-1)d]

let n=1

S1= 3n2+6n= 3(1)^2+6(1)

= 3+6=9

S2= 3(2)^2+6(2)=3(4)+12=24

S2= a1+a2

24=9+a2

a2= 15

A.P---->9,15.....

d= a2-a1= 15-9=6

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