Question

In a G.P. of even number of terms, the sum of all terms is five times the sum of the odd terms. The common ratio of the G.P. is 

पर्याय

• (a) \[- \frac{4}{5}\]

• (b) \[\frac{1}{5}\] 

• (b) \[\frac{1}{5}\] 

• (c) 4 

• (d) none of these 

Answer 1

(c) 4 

\[\text{ Let there be 2n terms in a G . P }. \]
\[\text{ Let a be the first term and r be the common ratio } . \]
\[ \because S_{2n} = 5\left( S_{\text{ odd terms }} \right)\]
\[ \Rightarrow \frac{a\left( r^{2n} - 1 \right)}{\left( r - 1 \right)} = 5\left( a + a r^2 + a r^4 + a r^6 + . . . a r^\left( 2n - 1 \right) \right)\]
\[ \Rightarrow \frac{a\left( r^{2n} - 1 \right)}{\left( r - 1 \right)} = 5\left( \frac{a\left( \left( r^2 \right)^n - 1 \right)}{\left( r^2 - 1 \right)} \right)\]
\[ \Rightarrow \frac{\left( r^{2n} - 1 \right)}{\left( r - 1 \right)} = 5\frac{\left( \left( r^2 \right)^n - 1 \right)}{\left( r^2 - 1 \right)}\]
\[ \Rightarrow \frac{\left( \left( r^n \right)^2 - 1^2 \right)}{\left( r - 1 \right)} = 5\frac{\left( \left( r^n \right)^2 - 1^2 \right)}{\left( r^2 - 1 \right)}\]
\[ \Rightarrow \frac{\left( r^n - 1 \right)\left( r^n + 1 \right)}{\left( r - 1 \right)} = 5\frac{\left( r^n - 1 \right)\left( r^n + 1 \right)}{\left( r - 1 \right)\left( r + 1 \right)}\]
\[ \Rightarrow \left( r^n - 1 \right)\left( r^n + 1 \right)\left( r - 1 \right)\left( r + 1 \right) - 5\left( r - 1 \right)\left( r^n - 1 \right)\left( r^n + 1 \right) = 0\]
\[ \Rightarrow \left( r^n - 1 \right)\left( r^n + 1 \right)\left( r - 1 \right)\left( r + 1 - 5 \right) = 0\]
\[\text{ But, r = 1 or - 1 is not possible }. \]
\[ \therefore r = 4\]
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