# Find the sum to n terms of the series 3 x 8 + 6 x 11 + 9 x 14 + ....

**Solution:**

The given series is 3 x 8 + 6 x 11 + 9 x 14 + .... n terms.

a_{n} = (n^{th} term of 3, 6, 9 ....) x (n^{th} term of 8, 11, 14 ....)

= (3n)(3n + 5)

= 9n^{2} + 15n

Therefore,

S_{n} = ∑^{n}_{k = 1}(a)_{k}

= ∑^{n}_{k = 1}(9k ^{2} + 15k)

= 9∑^{n}_{k = 1}(k)^{2} + 15∑^{n}_{k = 1}(k)

= 9 x [n (n + 1)(2n + 1)]/6 + 15 x [n (n + 1)]/2

= [3n (n + 1)(2n +1)]/2 + [15n (n + 1)]/2

= 3n (n + 1)/2 x (2n + 1 + 5)

= 3n (n + 1)/2 x (2n + 6)

= 3n (n + 1)(n + 3)

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.4 Question 6

## Find the sum to n terms of the series 3 x 8 + 6 x 11 + 9 x 14 + ....

**Summary:**

Therefore the sum to n terms of the series 3 x 8 + 6 x 11 + 9 x 14 + .... up to n terms is 3n (n + 1)(n + 3) using S_{n} = ∑^{n}_{k = 1}(a)_{k}