A series whose $n^{th}$ term is $\left( {\frac{n}{x}} \right) + y,$ the sum of $r$ terms will be
$\left\{ {\frac{{r(r + 1)}}{{2x}}} \right\} + ry$
$\left\{ {\frac{{r(r - 1)}}{{2x}}} \right\}$
$\left\{ {\frac{{r(r - 1)}}{{2x}}} \right\} - ry$
$\left\{ {\frac{{r(r + 1)}}{{2y}}} \right\} - rx$
Suppose we have an arithmetic progression $a_1, a_2, \ldots a_n, \ldots$ with $a_1=1, a_2-a_1=5$. The median of the finite sequence $a_1, a_2, \ldots, a_k$, where $a_k \leq 2021$ and $a_{k+1} > 2021$ is
For a series $S = 1 -2 + 3\, -\, 4 … n$ terms,
Statement $-1$ : Sum of series always dependent on the value of $n$ , i.e. whether it is even or odd.
Statement $-2$ : Sum of series is $-\frac {n}{2}$ when value of $n$ is any even integer
The number of common terms in the progressions $4,9,14,19, \ldots \ldots$, up to $25^{\text {th }}$ term and $3,6,9,12$, up to $37^{\text {th }}$ term is :
What is the sum of all two digit numbers which give a remainder of $4$ when divided by $6$ ?
Let $S_n$ be the sum to n-terms of an arithmetic progression $3,7,11, \ldots \ldots$. . If $40<\left(\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}\right)<42$, then $\mathrm{n}$ equals