If the pth term of an …

Let A be the first term and D be the common difference of the given AP. Then,
T_p = a {tex}\Rightarrow{/tex} A + (p - 1) D = a ...(i)
and T_q = b {tex}\Rightarrow{/tex} A + (q - 1) D = b ...(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(p - q) D = a - b {tex}\Rightarrow{/tex} D = {tex}\frac { a - b } { p - q }{/tex} ...(iii)
On adding Eqs. (i) and (ii), we get
2A + (p + q - 2) D = a + b
{tex}\Rightarrow{/tex} 2A + pD + qD - 2D = a + b
{tex}\Rightarrow{/tex} 2A + pD  + qD - D = a + b + D
{tex}\Rightarrow{/tex} 2A + (p + q - 1) D = a + b + D
2A + (p + q - 1) D = a + b + {tex}\left( \frac { a - b } { p - q } \right){/tex} [from Eq. (iii)] ...(iv)
Now, S_p+q = {tex}\frac { p + q } { 2 }{/tex} [2A + (p + q - 1) D]
= {tex}\frac { p + q } { 2 }{/tex} [a+ b + {tex}\frac { a - b } { p - q }{/tex}] [from Eq. (iv)]
Hence proved.