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`

Options

A. `(2m+1):(2n-1),`
B. `m:n`
C. `(2m-1):(2n-1)`
D. None of these

Explanation

To solve the problem, we need to show that the ratio of the mth and nth terms of an arithmetic progression (A.P.) is \((2m-1):(2n-1)\) given that the ratio of the sum of the first \(m\) and \(n\) terms is \(m^2:n^2\).

### Step-by-Step Solution:

1. **Understanding the Sum of Terms in A.P.**:
The sum of the first \(m\) terms of an A.P. is given by:
\[
S_m = \frac{m}{2} \left(2a + (m-1)d\right)
...