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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 49 of 106
Marks: +1, -0
Given a G.P. with a = 729 and 7th term 64, determine S7S_7.
Solution:  
Let a be the first term and the common ratio be r.
We have a = 729, a7a_7 = 64 ⇒ ar6ar^6 = 64
⇒ 729r6729r^6 = 64 ⇒ r6r^6 = 64729\frac{64}{729} ⇒ r6r^6 = (23)6\left(\frac{2}{3}\right)^6 ⇒ r = 23\frac{2}{3}
∴ S7S_7 = a(1−r7)1−r\frac{a(1-r^7)}{1-r} = 729 [1−(23)7]1−23\frac{\left[1-\left(\frac{2}{3}\right)^7\right]}{1-\frac{2}{3}} = 729 [1−1282187]13\frac{\left[1-\frac{128}{2187}\right]}{\frac{1}{3}} = 2187 [20592187]\left[\frac{2059}{2187}\right] = 2059
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