Question

If the sum of first 6 term is 9 times to the sum of first 3 terms of the same G.P., then the common ratio of the series will be.

• A. $- 2$
• B. 2
• C. 1
• D. ½

Answer 1

Answer: (B)

2

Answer 2

Under given conditions, we get

$\Rightarrow$ $\frac { a \left( r ^ { 6 } - 1 \right) } { ( r - 1 ) } = 9 \cdot \frac { a \left( r ^ { 3 } - 1 \right) } { ( r - 1 ) }$ $( \because r > 1 )$

$\Rightarrow$ $r ^ { 6 } - 1 = 9 r ^ { 3 } - 9$ $\Rightarrow$ $\left( r ^ { 3 } \right) ^ { 2 } - 9 \left( r ^ { 3 } \right) + 8 = 0$

$\Rightarrow$ $\left( r ^ { 3 } - 1 \right) \left( r ^ { 3 } - 8 \right) = 0$

$\Rightarrow$ $r = 1 , \omega , \omega ^ { 2 }$ and $r = 2$. But $r = 1 , \omega , \omega ^ { 2 }$ cannot satisfy the given condition. Hence $r = 2$.