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Question

The ratio of the sums of m and n terms of an A.P. is m2:n2. Show that the ratio of mth and nth term is (2m - 1) : (2n - 1).

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Solution

Let a be the first term and d be the common difference of the given A.P. Then, the sum of m and n terms are given by
Sm=m2[2a+(m1)d], and Sn=n2[2a+(n1)d]

Then,
SmSn=m2n2

m2[2a+(m1)d]n2[2a+(n1)d]=m2n2

2a+(m1)d2a+(n1)d=mn

[2a+(m1)d]n=[2a+(n1)d]m
2a(nm)=d[(n1)m(m1)n]
2a(nm)=d(nm)
d=2a

Therefore,

TmTn=a+(m1)da+(n1)d=a+(m1)2aa+(n1)2a=2m12n1

Hence proved

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