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

`S_(m)/(S_(n))=((m)/(2)[2a+(m-1)d])/((n)/(2)[2a+(n-a)d])=(m^(2))/(n^(2))`
`rArr (2a+(m-1)d)/(2a+(n-1)d)=(m)/(n)`
`rArr (a+((m-1)/(2))d)/(a+((n-1)/(2))d)=(m)/(n) " " ...(1)`
`"We want," (T_(m))/(T_(n)) "i.e.," (a+(m-1)d)/(a+(n-1)d)`
`"So, we replace " (m-1)/(2) " as " m-1`
`" i.e., " m-1 " as " 2(m-1)`
`rArr m-1 " as " 2m-2`
`rArr m " as " 2m-2+1`
i.e., replace m by 2m-1
Similarly, replace n by 2n-1 in eq. (1), we get
`:. (a+(m-1)d)/(a+(n-1)d)=(2m-1)/(2n-1)`
i.e., `(T_(m))/(T_(n)) = (2m-1)/(2n-1) " " ` Hence Proved.