If the sum of first $n$ terms of an $A.P.$ is $cn(n -1)$ , where $c \neq 0$ , then sum of the squares of these terms is
$c^2n^2(n+1)^2$
$\frac{2}{3}c^2n(n-1)(2n-1)$
$\frac{2}{3}c^2n(n+1)(2n+1)$
$\frac{c^2 n^2}{3}(n+1)^2$
Let the coefficients of the middle terms in the expansion of $\left(\frac{1}{\sqrt{6}}+\beta x\right)^{4},(1-3 \beta x)^{2}$ and $\left(1-\frac{\beta}{2} x\right)^{6}, \beta>0$, respectively form the first three terms of an $A.P.$ If $d$ is the common difference of this $A.P.$, then $50-\frac{2 d}{\beta^{2}}$ is equal to.
Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3}, \ldots$ be an $A.P.$ If $\frac{a_{1}+a_{2}+\ldots+a_{10}}{a_{1}+a_{2}+\ldots+a_{p}}=\frac{100}{p^{2}}, p \neq 10$, then $\frac{a_{11}}{a_{10}}$ is equal to :
Let ${a_1},{a_2},.......,{a_{30}}$ be an $A.P.$, $S = \sum\limits_{i = 1}^{30} {{a_i}} $ and $T = \sum\limits_{i = 1}^{15} {{a_{2i - 1}}} $.If ${a_5} = 27$ and $S - 2T = 75$ , then $a_{10}$ is equal to