If the sum of first $n$ terms of an $A.P.$ is $c n^2$, then the sum of squares of these $n$ terms is
$\frac{n\left(4 n^2-1\right) c^2}{6}$
$\frac{n\left(4 n^2+1\right) c^2}{3}$
$\frac{n\left(4 n^2-1\right) c^2}{3}$
$\frac{n\left(4 n^2+1\right) c^2}{6}$
The ratio of sum of $m$ and $n$ terms of an $A.P.$ is ${m^2}:{n^2}$, then the ratio of ${m^{th}}$ and ${n^{th}}$ term will be
Let ${a_1},{a_2},\;.\;.\;.\;.,{a_{49}}$ be in $A.P.$ such that $\mathop \sum \limits_{k = 0}^{12} {a_{4k + 1}} = 416$ and ${a_9} + {a_{43}} = 66$. If $a_1^2 + a_2^2 + \ldots + a_{17}^2 = 140m,$ then $m = \;\;..\;.\;.\;.\;$
If $a_1, a_2, a_3, …….$ are in $A.P.$ such that $a_1 + a_7 + a_{16} = 40$, then the sum of the first $15$ terms of this $A.P.$ is
If $p$ times the ${p^{th}}$ term of an $A.P.$ is equal to $q$ times the ${q^{th}}$ term of an $A.P.$, then ${(p + q)^{th}}$ term is
In $\Delta ABC$, if $a, b, c$ are in $A.P.$ (with usual notations), identify the incorrect statements -