Question

Show that the sum of an arithmetic progression whose first term is 'a' the second term is 'b' and last term is'c' is (a+c)(b+c-2a) /2(b-a)

Answer 1

Hey dear friend

last term=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

So, sum of n terms
By the formula,= Sn=n/2[2a+(n-1)d]

Sn= 1/2[(c-a)/(b-a)+1] [2a+(c-a)/b-a)(b-a)]
Sn=1/2[(c-a+b-a)/(b-a)] [2a+c-a]
Sn=[(c+b-2a)(a+c)/2(b-a)]

perfectly fine pure correct answer
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