Let $s _1, s _2, s _3, \ldots \ldots, s _{10}$ respectively be the sum to 12 terms of 10 A.P.s whose first terms are $1,2,3, \ldots, 10$ and the common differences are $1,3,5, \ldots, 19$ respectively. Then $\sum \limits_{i=1}^{10} s _{ i }$ is equal to

The $p^{\text {th }}, q^{\text {th }}$ and $r^{\text {th }}$ terms of an $A.P.$ are $a, b, c,$ respectively. Show that $(q-r) a+(r-p) b+(p-q) c=0$

If ${a^2},\,{b^2},\,{c^2}$ be in $A.P.$, then $\frac{a}{{b + c}},\,\frac{b}{{c + a}},\,\frac{c}{{a + b}}$ will be in

Given that $n$ A.M.'s are inserted between two sets of numbers $a,\;2b$and $2a,\;b$, where $a,\;b \in R$. Suppose further that ${m^{th}}$ mean between these sets of numbers is same, then the ratio $a:b$ equals

The sums of $n$ terms of three $A.P.'s$ whose first term is $1$ and common differences are $1, 2, 3$ are ${S_1},\;{S_2},\;{S_3}$ respectively. The true relation is

Let ${S_1},{S_2},......,{S_{101}}$ be the consecutive terms of an $A.P$ . If $\frac{1}{{{S_1}{S_2}}} + \frac{1}{{{S_2}{S_3}}} + .... + \frac{1}{{{S_{100}}{S_{101}}}} = \frac{1}{6}$ and ${S_1} + {S_{101}} = 50$ , then $\left| {{S_1} - {S_{101}}} \right|$ is equal to