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)

= 9n² + 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)

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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