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Question
In an AP. Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to
Options
0
−(p + q)
p + q
pq
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Solution
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) `
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