Question

If the sum of the first m terms of an AP is n and the sum of the first n terms is m, then show that the sum of its (m+n)th term is -(m+n).

Answer 1

Given :-

• The Sum of the first m terms of an AP is n.

• The Sum of its first n terms is m.

To show :-

• The sum of its (m+n) terms is -(m+n)

Solution :-

Let a be the first term and d the common difference of the given AP.

then,

$\rm\:S_m=n$

$\rm\longrightarrow\:S_m=\dfrac{m}{2}\:[2a+(m-1)d]=n$

$\rm\longrightarrow\:S_m=2am+m(m-1)d=2n\:........(i)$

And,

$\rm\:S_n=m$

$\rm\longrightarrow\:S_n=\dfrac{n}{2}\:[2a+(n-1)d]=m$

$\rm\longrightarrow\:S_n=2an+n(n-1)d=2m\:........(ii)$

❍ Subtracting eq (ii) from eq (i) ,we get

$\rm\:2a(m-n)+[({m}^{2}-{n}^{2})-(m-n)]d=2(n-m)$

$\rm\longrightarrow\:(m-n)[2a+(m+n-1)d]=2(n-m)$

$\rm\longrightarrow\:2a+(m+n-1)d=-2\:............(iii)$

❍ sum of the first (m+n) terms of the given AP

$\rm\longrightarrow\:\dfrac{(m+n)}{2}.[2a+(m+n-1)d]$

$\rm\longrightarrow\:\dfrac{(m+n)}{2}.(-2)$

$\rm\longrightarrow\:-(m+n)\:\:\:\:\:\:\:\:\:[using\:eq\:(iii)]$

Hence,the sum of first (m+n) terms of the given AP is -(m+n).