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

Ans:  Let:

First term of given A.P. = a,

Common difference = d,

Then, the sums of m and n terms are given by



Then, \frac{S_{m}}{S_{m}}=\frac{m^{2}}{n^{2}}

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

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

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

\implies 2a(n-m)=d\{(n-1)m-(m-1)n\}

\implies 2a(n-m)=d(n-m)

\implies d=2a


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