Question

If the sum of first p terms of an AP is equal to the sum of first q terms, then the sum of first (p+q)
terms i.e., Sp+q is
​

Answer 1

Answer:

$S_{p+q}$   = 0

Step-by-step explanation:

Given sum of first p terms of AP = sum of first q terms of it

ie,   $S_{p}$ = $S_{q}$   ;   $S_{p+q}$ = ?

so,

    $S_{p}$ = $S_{q}$

=> $\frac{p}{2} * [ 2a + (p-1)d ]$  = $\frac{q}{2} * [ 2a + (q-1)d ]$

=>  $p * [ 2a + (p-1)d ]$  = $q* [ 2a + (q-1)d ]$

=> 2ap + p(p-1)d = 2aq + q(q-1)d

=> 2ap - 2aq = q(q-1)d - p(p-1)d

=> 2a (p-q) = d [ q(q-1) - p(p-1) ]

=> 2a (p-q) = d [ q² - q - p² + p) ]

=> 2a (p-q) = - d [ p² - q² - p + q ]

=> 2a (p-q) = - d [ (p² - q²) - ( p - q) ]

=> 2a (p-q) = - d [ (p - q)(p+q) - ( p - q) ]

=> 2a (p-q) = - (p - q)d [ (p+q) - 1 ]

=> 2a  = - d [ (p+q) - 1 ]

=> 2a + (p+q - 1)d = 0  ------------- (1)

Now (by formula) , $S_{p+q}$ = $\frac{p+q}{2} * [ 2a + (p+q-1)d ]$

                                       = $\frac{p+q}{2} *0$   by (1)

                         => $S_{p+q}$   = 0

                         => Sum of first (p+q) terms = 0