If the p th term of an AP is a and q th term is b then show that the sum of its first (p+q) terms is p+q/2[a+b+a-b/p+q][2p+1].
Let 'α' be the first term, and 'd' the common difference of the A.P..
T(n) = α + (n-1)d
S(n) = (n/2) [ 2α + (n-1) d ] ................. (1)
Now, given that : T(p) = a and T(q) = b.
∴ α + (p-1)d = a .......... (2)
α + (q-1)d = b .......... (3)
Subtracting (3) from (2),
(p-q)d = (a-b)
∴ d = (a-b) / (p-q) ....................... (4)
Adding (2) and (3),
2α + (p+q-2)d = a + b
∴ [ 2α + (p+q-1)d ] - d = a + b
∴ [ 2α + (p+q-1)d ] = (a+b) + d
∴ from (4),
[ 2α + (p+q-1)d ] = (a+b) + (a-b)/(p-q) .......... (5)
Putting n = (p+q) in(1), and then using (5),
S(p+q) = [ (p+q)/2 ]· [ 2α + (p+q-1)d ]
= [ (p+q)/2 ]· [ (a+b) + (a-b)/(p-q) ]
We hope that clarifies your query.
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