# The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P

**Solution:**

Let the G.P be a, ar, ar^{2}, ar^{3}, ....

According to the given question

a + ar + ar^{2} = 16

and ar^{3} + ar^{4} + ar^{5} = 128

Therefore,

a (1 + r + r ^{2}) = 16 ....(1)

ar^{3} (1 + r + r ^{2}) = 128 ....(2)

Dividing (2) by (1), we obtain

⇒ [ar^{3} (1+ r + r^{2})] / [a (1+ r + r^{2})]

= 128/16

⇒ r^{3} = 8

⇒ r = 2

Substituting r = 2 in (1) , we obtain

⇒ a (1 + 2 + 4) = 16

⇒ a (7) = 16

⇒ a = 16/7

Hence,

S_{n} = a (1 - r^{n})/(1 - r)

= 16/7 (2^{n} - 1)/(2 - 1)

= 16/7 (2^{n} - 1)

Thus, the first term, a = 16/7, common ratio, r = 2 and sum to n terms, S_{n} = 16/7 (2^{n} - 1)

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 14

## The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P

**Summary:**

It is known that the first three terms of the G.P was 15 and the sum of the next three terms is 128, the first term is 16/7, the common ratio is 2 and the sum of n terms is 16/7 (2^{n} - 1)

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