# 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 - Mathematics

MCQ

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

#### Options

• (a) $- \frac{4}{5}$

• (b) $\frac{1}{5}$

• (b) $\frac{1}{5}$

• (c) 4

• (d) none of these

#### Solution

(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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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Q 20 | Page 58