Gujarati
8. Sequences and Series
hard

Suppose the sum of the first $m$ terms of an arithmetic progression is $n$ and the sum of its first $n$ terms is $m$, where $m \neq n$. Then, the sum of the first $(m+n)$ terms of the arithmetic progression is

A

$1-m n$

B

$m n-5$

C

$-(m+n)$

D

$m+n$

(KVPY-2018)

Solution

(c)

Given, $S_m=n$ and $S_n=m$

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

$S_n=\frac{n}{2}(2 a+(n-1) d)=m \text { (i) }$

On subtracting Eq.$(ii)$ from Eq.$(i)$, we get

$(m-n) a+(m-n)(m+n-1) \frac{d}{2}$

$=-(m-n)$

$2 a+(m+n-1) d=-2 \quad[m \neq n]$

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

$=\frac{m+n}{2}(-2)=-(m+n)$

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.