Given sum of the first $n$ terms of an $A.P.$ is $2n + 3n^2.$ Another $A.P.$ is formed with the same first term and double of the common difference, the sum of $n$ terms of the new $A.P.$ is
$n + 4n^2$
$6n^2 - n$
$n^2 + 4n$
$3n + 2n^2$
In an $A.P.,$ the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that $20^{th}$ term is $-112$
Let $S_{1}$ be the sum of first $2 n$ terms of an arithmetic progression. Let, $S_{2}$ be the sum of first $4n$ terms of the same arithmetic progression. If $\left( S _{2}- S _{1}\right)$ is $1000,$ then the sum of the first $6 n$ terms of the arithmetic progression is equal to:
The value of $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} $ is
Find the sum of all numbers between $200$ and $400$ which are divisible by $7.$
If $n$ be odd or even, then the sum of $n$ terms of the series $1 - 2 + $ $3 - $$4 + 5 - 6 + ......$ will be