Let the positive numbers $a _1, a _2, a _3, a _4$ and $a _5$ be in a G.P. Let their mean and variance be $\frac{31}{10}$ and $\frac{ m }{ n }$ respectively, where $m$ and $n$ are co-prime. If the mean of their reciprocals is $\frac{31}{40}$ and $a_3+a_4+a_5=14$, then $m + n$ is equal to $.........$.
$210$
$212$
$213$
$211$
If $n$ geometric means be inserted between $a$ and $b$ then the ${n^{th}}$ geometric mean will be
An $A.P.$, a $G.P.$ and a $H.P.$ have the same first and last terms and the same odd number of terms. The middle terms of the three series are in
If five $G.M.’s$ are inserted between $486$ and $2/3$ then fourth $G.M.$ will be
If $a,\,b,\,c$ are in $A.P.$ and ${a^2},\,{b^2},{c^2}$ are in $H.P.$, then
If $a _{1}(>0), a _{2}, a _{3}, a _{4}, a _{5}$ are in a G.P., $a _{2}+ a _{4}=2 a _{3}+1$ and $3 a _{2}+ a _{3}=2 a _{4}$, then $a _{2}+ a _{4}+2 a _{5}$ is equal to