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

Explanation

Let a be the first term and d the common difference of the given A.P. Then, the sums of m and n terms are given by
`S_(m)=m/2[2a+(m-1)d]`
and `S_(n)=n/2[2a+(n-1)d]`
Given, `(S_(m))/(S_(n))=m^(2)/(n^(2))`
or `(2a+(m-1)d)/(2a+(n-1)d)=m/n`
replacing m by 2m-1 and n by 2n-1, we have
`(2a+(2m-1-1)d)/(2a+(2n-1-1)d)=(2m-1)/(2n-1)`
or `(a+(m-1)d)/(a+(n-1)d)=(2m-1)/(2n-1)`
`rArr(T_(m))/(T_(n))=(2m-1)/(2n-1)`