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 the sums of first $n$ even numbers and $n$ odd numbers will be
Let $a, b, c, d, e$ be natural numbers in an arithmetic progression such that $a+b+c+d+e$ is the cube of an integer and $b+c+d$ is square of an integer. The least possible value of the number of digits of $c$ is
$8^{th}$ term of the series $2\sqrt 2 + \sqrt 2 + 0 + .....$ will be
If $a _{1}, a _{2}, a _{3} \ldots$ and $b _{1}, b _{2}, b _{3} \ldots$ are $A.P.$ and $a_{1}=2, a_{10}=3, a_{1} b_{1}=1=a_{10} b_{10}$ then $a_{4} b_{4}$ is equal to
The value of $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} $ is