Let the sequence ${a_1},{a_2},{a_3},.............{a_{2n}}$ form an $A.P. $ Then $a_1^2 - a_2^2 + a_3^3 - ......... + a_{2n - 1}^2 - a_{2n}^2 = $
If the sum of $\mathrm{n}$ terms of an $\mathrm{A.P.}$ is $n P+\frac{1}{2} n(n-1) Q,$ where $\mathrm{P}$ and $\mathrm{Q}$ are constants, find the common difference.
If the sum of first $11$ terms of an $A.P.$, $a_{1} a_{2}, a_{3}, \ldots$is $0\left(\mathrm{a}_{1} \neq 0\right),$ then the sum of the $A.P.$, $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1},$ where $k$ is equal to
If $a,\;b,\;c$ are in $A.P.$, then $\frac{1}{{bc}},\;\frac{1}{{ca}},\;\frac{1}{{ab}}$ will be in
Let $S_n$ and $s_n$ deontes the sum of first $n$ terms of two different $A.P$. for which $\frac{{{s_n}}}{{{S_n}}} = \frac{{3n - 13}}{{7n + 13}}$ then $\frac{{{s_n}}}{{{S_{2n}}}}$