If S_1, S_2, S_3, dots, S_m are the sums of n | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The sum of $n$ terms of an $A.P.$ is given by $S = \frac{n}{2}[2a + (n - 1)d]$.For the $k$-th $A.P.$,the first term $a_k = k$ and the common diffe

Vedclass · Accepted Answer

$\frac{1}{2}mn(mn + 1)$

Vedclass · Answer

(A) The sum of $n$ terms of an $A.P.$ is given by $S = \frac{n}{2}[2a + (n - 1)d]$.For the $k$-th $A.P.$,the first term $a_k = k$ and the common difference $d_k = 2k - 1$.Thus,$S_k = \frac{n}{2}[2k + (n - 1)(2k - 1)] = \frac{n}{2}[2k + 2kn - n - 2k + 1] = \frac{n}{2}[2kn - n + 1]$.Now,we need to find the sum $\sum_{k=1}^{m} S_k = \sum_{k=1}^{m} \frac{n}{2}[2kn - n + 1]$.$= \frac{n}{2} [2n \sum_{k=1}^{m} k + \sum_{k=1}^{m} (1 - n)]$.$= \frac{n}{2} [2n \frac{m(m+1)}{2} + m(1 - n)]$.$= \frac{n}{2} [nm(m+1) + m - mn] = \frac{n}{2} [nm^2 + nm + m - mn] = \frac{n}{2} [nm^2 + m] = \frac{mn(mn + 1)}{2}$.

If $S_1, S_2, S_3, \dots, S_m$ are the sums of $n$ terms of $m$ arithmetic progressions $(A.P.)$ whose first terms are $1, 2, 3, \dots, m$ and common differences are $1, 3, 5, \dots, 2m - 1$ respectively,then $S_1 + S_2 + S_3 + \dots + S_m = $

Similar Questions

Four numbers are in arithmetic progression. The sum of the first and last term is $8$ and the product of both middle terms is $15$. The least number of the series is

Let ${S_n}$ denote the sum of $n$ terms of an $A.P.$ If ${S_{2n}} = 3{S_n}$,then the ratio $\frac{{{S_{3n}}}}{{{S_n}}}$ is equal to:

Let $f: R \rightarrow R$ be such that for all $x \in R$,$(2^{1+x}+2^{1-x})$,$f(x)$,and $(3^x+3^{-x})$ are in $A.P.$. Then,the minimum value of $f(x)$ is:

If $S_n = \frac{n(n + 1)(n + 2)}{6}$,then $\sum_{n = 1}^\infty \frac{1}{t_n} = $

If the sum of the first $6$ terms is $9$ times the sum of the first $3$ terms of the same $G.P.$,then the common ratio of the series will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module