Sum of the first $p, q$ and $r$ terms of an $A.P.$ are $a, b$ and $c,$ respectively. Prove that $\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$
If ${A_1},\,{A_2}$ be two arithmetic means between $\frac{1}{3}$ and $\frac{1}{{24}}$ , then their values are
If $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are in $H.P.$, then
The ratio of the sums of first $n$ even numbers and $n$ odd numbers will be
If ${m^{th}}$ terms of the series $63 + 65 + 67 + 69 + .........$ and $3 + 10 + 17 + 24 + ......$ be equal, then $m = $