Find the sum to $n$ terms of the $A.P.,$ whose $k^{\text {th }}$ term is $5 k+1$

Vedclass pdf generator app on play store
Vedclass iOS app on app store

It is given that the $k^{\text {th }}$ term of the $A.P.$ is $5 k+1$

$k^{\text {th }}$ term $=a_{k}+(k-1) d$

$\therefore a+(k-1) d=5 k+1$

$a+k d-d=5 k+1$

$\therefore$ Comparing the coefficient of $k ,$ we obtain $d=5$

$\Rightarrow a-d=1$

$\Rightarrow a-5=1$

$\Rightarrow a=6$

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{n}{2}[2(6)+(n-1)(5)]$

$=\frac{n}{2}[12+5 n-5]$

$=\frac{n}{2}[5 n+7]$

Similar Questions

Let the sum of the first $n$ terms of a non-constant $A.P., a_1, a_2, a_3, ……$ be $50\,n\, + \,\frac{{n\,(n\, - 7)}}{2}A,$ where $A$ is a constant. If $d$ is the common difference of this $A.P.,$ then the ordered pair $(d,a_{50})$ is equal to

  • [JEE MAIN 2019]

Let $a_{1}, a_{2} \ldots, a_{n}$ be a given $A.P.$ whose common difference is an integer and $S _{ n }= a _{1}+ a _{2}+\ldots+ a _{ n }$ If $a_{1}=1, a_{n}=300$ and $15 \leq n \leq 50,$ then the ordered pair $\left( S _{ n -4}, a _{ n -4}\right)$ is equal to

  • [JEE MAIN 2020]

Let $a_n, n \geq 1$, be an arithmetic progression with first term $2$ and common difference $4$ . Let $M_n$ be the average of the first $n$ terms. Then the sum $\sum \limits_{n=1}^{10} M_n$ is

  • [KVPY 2019]

The sum of all the elements of the set $\{\alpha \in\{1,2, \ldots, 100\}: \operatorname{HCF}(\alpha, 24)=1\}$ is

  • [JEE MAIN 2022]

If $\frac{{{a^{n + 1}} + {b^{n + 1}}}}{{{a^n} + {b^n}}}$ be the $A.M.$ of $a$ and $b$, then $n=$