- Home
- Standard 11
- Mathematics
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
