Question

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

Answer 1

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

= 3 + 6n

Answer 2

Answer:

AP means arithmetic progression which is a series having the common difference between any two consecutive terms equal.

Given,

Sum of first n terms = 3n^2 +6n

to find : nth term of this AP

Explanation:

let the value of n be 1

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

S1 = 3+6=9

S1 = a1 = 9

sum of first term = first term

now let the value of n be 2

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

S2 = 12+12=24

S2 = a1 + a2

sum of two terms = sum of first and second terms

24= 9 +a2

24-9 = a2

15= a2

d= common difference = a2-a1

d= 15-9

d= 6

an = a+(n-1)d

an = 9 + (n-1)(6)

an = 9+6n-6

an =3+6n