The first, second and the last term of an AP is 'a', 'b' and 'c' respectively. Prove that the sum is
(a + c)(b+c-2a)/
2(b-a)
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Explanation:
a,b....c
First term=a
Common diff.=b-a
Tn=a+(n-1)d
c=a+(n-1)b-a
c-a=(n-1)b-a
(c-a)/(b-a)=n-1
((c-a)/(b-a)+1)=n
Taking L.C.M.
(c-a+b-a)/b-a=n
Therefore,
(b+c-2a)/b-a = n
Sn=n/2(a+l) where l=last term
=(b+c-2a)/2(b-a) * (a+c)
Hence, proved
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