If $$\left( {1 + 3 + 5 + ..... + p} \right) + \left( {1 + 3 + 5 + ..... + q} \right) = \left( {1 + 3 + 5 + ..... + r} \right)$$              where each set of parentheses contains the sum of consecutive odd integers as shown, what is the smallest possible value of $$\left( {p + q + r} \right){\text{where }}p > 6?$$

Solution :
Since $$n^{th}$$ term of A.P. $$ = a + \left( {n - 1} \right)d$$
$$\therefore p = 1 + \left( {n - 1} \right)2$$
( ∵ First term $$= a = 1$$  and common difference $$= d = 2$$  )
$$\eqalign{ & \Rightarrow n = \frac{{p + 1}}{2} \cr & \therefore \left( {1 + 3 + 5 + ..... + p} \right) + \left( {1 + 3 + 5 + ..... + q} \right) = \left( {1 + 3 + 5 + ..... + r} \right) \cr & \Rightarrow \frac{{\frac{{p + 1}}{2}}}{2}\left[ {2 \times 1 + \left( {\frac{{p + 1}}{2} - 1} \right)2} \right] + \frac{{\left( {\frac{{q + 1}}{2}} \right)}}{2}\left[ {2 \times 1 + \left( {\frac{{q + 1}}{2} - 1} \right)2} \right] \cr & = \frac{{r + 1}}{4}\left[ {2 \times 1 + \left( {\frac{{r + 1}}{2} - 1} \right)2} \right] \cr & \Rightarrow \frac{{p + 1}}{4}\left[ {2 + \left( {p - 1} \right)} \right] + \frac{{q + 1}}{4}\left[ {2 + \left( {q - 1} \right)} \right] \cr & = \frac{{r + 1}}{4}\left[ {2 + r - 1} \right] \cr & \Rightarrow {\left( {p + 1} \right)^2} + {\left( {q + 1} \right)^2} = {\left( {r + 1} \right)^2} \cr} $$
This is the possible only when $$p = 7, q = 5, r = 9$$
$$ \therefore p + q + r = 7 + 5 + 9 = 21$$