# Given a G.P with a = 729 and 7^{th} term 64, determine S_{7}

**Solution:**

It is given that

a = 729 and a_{7} = 64

Let r be the common ratio of the G.P.

It is known that Therefore, a = ar^{n }^{- 1}

⇒ a = ar^{7 }^{- }^{1} = ar^{6}

⇒ 64 = 729r^{6}

r^{6} = (2/3)^{6}

r = 2/3

Also,

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

= 729 (1 - (2/3)^{7})/(1 - 2/3)

= 3 x 729 (1 - (2/3)^{7})

(3)^{7} [(3)^{7} - (2)^{7}]/(3)^{7}

(3)^{7} - (2)^{7}

= 2187 - 128

= 2059

NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 15

## Given a G.P with a = 729 and 7^{th} term 64, determine S_{7}

**Summary:**

Given that the first term =729 and the 7th term is 64 we find out that the sum of seven numbers is 2059

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