The ratio of the sum of m and n terms of an A.P. is m^2:n^2

Q) If the sum of m terms of an A.P. is the same as the sum of its n terms, show that the sum of its (m+n) terms is zero.

Ans: Let

$a$ = First term of the given A.P. and

$d$ = Common difference .

Then,

$S_{m}=S_{n}$

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

$2am + m(m-1)d = 2an + n(n-1)d$

$2am + m(m-1)d - 2an - n(n-1)d = 0$

$2am - 2an + m(m-1)d - n(n-1)d = 0$

$2a(m-n)+\{m(m-1)-n(n-1)\}d=0$

$2a(m-n)+\{(m^{2}-n^{2})-(m-n)\}d=0$

$(m-n)\{2a+(m+n-1)d\}=0$

$2a+(m+n-1)d=0$ . $[(m-n)\neq 0]$

Now, $S_{m+n}=\frac{m+n}{2}\{2a+(m+n-1)d\}=\frac{m+n}{2}\times0=0$

Hence Proved!

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