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Question

If the sum of first m terms of an A.P. is n and the sum of its first n terms is m, then find the sum of its first (m+n) terms.

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Solution

Sum of p terms of an A.P. is given by =p2[2a+(p1)d]
Given that sum of first m terms of an A.P. is n, i.e.,
Sm=n
m2[2a+(m1)d]=n
2am+m2dmd=2n.....(1)
Also given that sum of first n terms of an A.P. is m, i.e.,
Sn=m
n2[2a+(n1)d]=m
2an+n2dnd=2m.....(2)
Subtracting eqn(2) from (1), we have
(2am+m2dmd)(2an+n2dnd)=2n2m
2am+m2dmd2ann2d+nd=2(nm)
2a(mn)+(m2n2)d(mn)d=2(nm)
(mn)(2a+(m+n)dd)=2(mn)
2a+((m+n)1)d=2
Now for sum of first (m+n) terms-
S(m+n)=m+n2[2a+((m+n)1)d]
S(m+n)=(m+n)2×(2)=(m+n)
Hence the sum of first (m+n) terms is (m+n).

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