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Question
The pth term of an A.P. is a and qth term is b. Prove that the sum of its (p + q) terms is `(p + q)/2[a + b + (a - b)/(p - q)]`.
Solution
Let A be the first term and D be the common difference of the A.P.
It is given that tp = a
⇒ A + (p – 1) D = a .....(1)
tq = b
⇒ A + (q – 1) D = b .....(2)
Subtracting (2) from (1), we get
(p – 1 – q + 1) D = a – b
⇒ D = `(a - b)/(p - q)` .....(3)
Adding (1) and (2), we get
2A + (p + q – 2) D = a + b
⇒ 2A + (p + q – 1) D = a + b + D
⇒ 2A + (p + q – 1) D = `a + b"+ (a - b)/(p - q)` ....(4)
Now Sp+q = `(p + q)/2 [2"A" + (p + q - 1) "D"]`
= `(p + q)/2[a + b + (a - b)/(p - q)]` ...[(Using ...(3) and (4)]
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