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`S_m`and`S_n ` be the sum of the first m and first n terms of the AP respectively.
Let, a be the first term and d be a common difference

`S_n/S_m`=`m ^2/n^2`

⇒ `n/2 [2a+(n−1)d]/m/2
[2a+(m−1)d]= (n /2/m /2)`

` ⇒ 2a+(n−1)d2a+(m−1)d
​ = m`



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