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,$\sum_{k=1}^{m} S_k = \sum_{k=1}^{m} \frac{n}{2}[2kn - n + 1] = \frac{n}{2} [2n \sum_{k=1}^{m} k - \sum_{k=1}^{m} (n - 1)]$.$= \frac{n}{2} [2n \frac{m(m+1)}{2} - m(n-1)] = \frac{n}{2} [nm(m+1) - m(n-1)]$.$= \frac{nm}{2} [n(m+1) - (n-1)] = \frac{nm}{2} [nm + n - n + 1] = \frac{1}{2}mn(mn + 1)$.

If $S_1, S_2, S_3, \dots, S_m$ are the sums of $n$ terms of $m$ $A.P.s$ 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

Let $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}, \dots, \frac{1}{x_n}$ ($x_i \neq 0$ for $i = 1, 2, \dots, n$) be in $A.P.$ such that $x_1 = 4$ and $x_{21} = 20$. If $n$ is the least positive integer for which $x_n > 50$,then $\sum_{i=1}^n \left( \frac{1}{x_i} \right)$ is equal to:

If the ratio of the sum of $n$ terms of two $A.P.s$ is $(7n + 1):(4n + 27)$,then the ratio of their $11^{th}$ terms is:

If the sides of a right-angled triangle are in an arithmetic progression,then they are in the ratio of.........

The sum of the numbers between $100$ and $1000$ which are divisible by $9$ is:

The sum of the series $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} + \dots$ to $9$ terms is

Vedclass Test Series

Exam Paper Generator

Online Exam Module