Question

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)​

Answer 1

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

Answer 2

Step-by-step explanation:

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