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≠n. Then, the sum of the first…

Question

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≠n. Then, the sum of the first (m+n) terms of the arithmetic progression is

MARKS Content Team · Accepted Answer

Correct answer: (C) -(m+n). Given, S_m=n and S_n=m S_m=m/2[2a+(m-1)d]=n...(i) S_n=n/2(2a+(n-1)d)=m...(ii) On subtracting Eq. (ii) from Eq. (i), we get (m-n)a+(m-n)(m+n-1)d/2 =-(m-n) ⇒ 2a+(m+n-1)d=-2 [m≠n] ∴ S_(m+n)=(m+n)/2(2a+(m+n-1)d) =(m+n)/2(-2)=-(m+n)

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

Solution

More Sequences and Series Questions