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).

Explanation

Let Sm​ and Sn​ 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

`(Sn)/(​Sm)​​=(n^2)/(m^2)​`


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

`⇒(2a+(m−1)d​)/(2a+(n−1)d)=m/n​`

`⇒n[2a+(m−1)d]=m[2a+(n−1)d]`

`⇒2an+mnd−nd=2am+mnd−nd`

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