If `S_1, S_2, S_3` be respectively the sums of `n ,2n ,3n` terms of a G.P., then prove that `(S_1)^2+(S_2)^2=S_1(S_2+S_3)` .

Explanation

To prove that \( S_1^2 + S_2^2 = S_1(S_2 + S_3) \), where \( S_1, S_2, S_3 \) are the sums of \( n, 2n, 3n \) terms of a geometric progression (G.P.), we can follow these steps:

### Step 1: Define the sums of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The sums of the first \( n \), \( 2n \), and \( 3n \) terms can be expressed as:

\[ S_1 = \frac{a(r^n - 1)}{r - 1} \]
...