Question

If $( p + q ) ^ { t h }$ term of a G.P. be $m$ and $( p - q ) ^ { t h }$ term be $n$, then the $p ^ { t h }$ term will be.

• A. $m / n$
• B. $\sqrt { m n }$
• C. $m n$
• D. 0

Answer 1

Answer: (B)

$\sqrt { m n }$

Answer 2

Given that $m = a r ^ { p + q - 1 }$ and $n = a r ^ { p - q - 1 }$

$r ^ { p + q - 1 - p + q + 1 } = \frac { m } { n } \Rightarrow r = \left( \frac { m } { n } \right) ^ { 1 / ( 2 q ) }$ and $a = \frac { m } { \left( \frac { m } { n } \right) ^ { ( p + q - 1 ) / 2 q } }$

Now $p ^ { t h }$term

$= a r ^ { p - 1 } = \frac { m } { \left( \frac { m } { n } \right) ^ { ( p + q - 1 ) / 2 q } } \left( \frac { m } { n } \right) ^ { ( p - 1 ) / 2 q }$

$= m \left( \frac { m } { n } \right) ^ { ( p - 1 ) / 2 q - ( p + q - 1 ) / ( 2 q ) } = m \left( \frac { m } { n } \right) ^ { - 1 / 2 } = m ^ { 1 - 1 / 2 } n ^ { 1 / 2 }$

$= m ^ { 1 / 2 } n ^ { 1 / 2 } = \sqrt { m n }$ .

Aliter : As we know each term in a G.P. is geometric mean of the terms equidistant from it. Here $( p + q ) ^ { t h }$ and $( p - q ) ^ { t h }$ terms are equidistant from $p ^ { t h }$ term at a distance of $q$. Therefore, $p ^ { t h }$ term will be G.M. of $( p + q ) ^ { t h }$ and $( p - q ) ^ { t h }$ $\sqrt { m n }$ .