If $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are in $H.P.$, then
${a^2}b,\,{c^2}a,\,{b^2}c$ are in $A.P.$
${a^2}b,\,{b^2}c,\,{c^2}a$ are in $H.P.$
${a^2}b,\,{b^2}c,\,{c^2}a$ are in $G.P.$
None of these
For any three positive real numbers $a,b,c$ ; $9\left( {25{a^2} + {b^2}} \right) + 25\left( {{c^2} - 3ac} \right) = 15b\left( {3a + c} \right)$ then
Insert five numbers between $8$ and $26$ such that resulting sequence is an $A.P.$
The value of $x$ satisfying ${\log _a}x + {\log _{\sqrt a }}x + {\log _{3\sqrt a }}x + .........{\log _{a\sqrt a }}x = \frac{{a + 1}}{2}$ will be
Let $a_1, a_2, a_3, \ldots, a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\sum_{i=1}^p a_i, 1 \leq p \leq 100$. For any integer $n$ with $1 \leq n \leq 20$, let $m=5 n$. If $\frac{S_{m m}}{S_n}$ does not depend on $n$, then $a_2$ is
Let $a_{1}, a_{2}, \ldots \ldots, a_{21}$ be an $A.P.$ such that $\sum_{n=1}^{20} \frac{1}{a_{n} a_{n+1}}=\frac{4}{9}$. If the sum of this AP is $189,$ then $a_{6} \mathrm{a}_{16}$ is equal to :