Question

In an AP. S_p = q, S_q = p and S_r denotes the sum of first r terms. Then, S_p+q is equal to

Options

• 0

• −(p + q)

•  p + q

• pq

Answer 1

In the given problem, we are given `S_p - q` and  `S_q = p`

We need to find  `S_( p +q)`

Now, as we know,

`S_n = n/2 [ 2a + (n-1) d]`

So,

`S_p = p/2 [2a + (p-1) d]`

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

2q = 2ap + p (p - 1)d                        .................(1) 

Similarly,

`S_q = q/2 [2a + (q - 1) d ]`

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

2p = 2aq  + q(q-1)d                          ......................(2)

Subtracting (2) from (1), we get

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

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

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

       - 2 = 2a + ( p + q - 1 ) d                           ..................(3) 

Now,

`S_(p + q) = ( p+q)/2 [2a + ( p+q - 1 ) d ]`

`S_(p+q) = ((p+q))/2 (-2)   `             ....(Using 3 ) 

`S_(p+q) = - (p+ q) `

Thus,  `S_(p+q) = - (p+ q) `