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, ar², ar³, ....

According to the given question

a + ar + ar² = 16

and ar³ + ar⁴ + ar⁵ = 128

Therefore,

a (1 + r + r ²) = 16 ....(1)

ar³ (1 + r + r ²) = 128 ....(2)

Dividing (2) by (1), we obtain

⇒ [ar³ (1+ r + r²)] / [a (1+ r + r²)]

= 128/16

⇒ r³ = 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 - rⁿ)/(1 - r)

= 16/7 (2ⁿ - 1)/(2 - 1)

= 16/7 (2ⁿ - 1)

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

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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 (2ⁿ - 1)