The $8^{\text {th }}$ common term of the series $S _1=3+7+11+15+19+\ldots . .$ ; $S _2=1+6+11+16+21+\ldots .$ is $.......$.

If twice the $11^{th}$ term of an $A.P.$ is equal to $7$ times of its $21^{st}$ term, then its $25^{th}$ term is equal to

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 ${m^{th}}$ terms of the series $63 + 65 + 67 + 69 + .........$ and $3 + 10 + 17 + 24 + ......$ be equal, then $m = $

If $\frac{1}{{b - c}},\;\frac{1}{{c - a}},\;\frac{1}{{a - b}}$ be consecutive terms of an $A.P.$, then ${(b - c)^2},\;{(c - a)^2},\;{(a - b)^2}$ will be in

The $20^{\text {th }}$ term from the end of the progression $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots .,-129 \frac{1}{4}$ is :-