If first term ,second , and last term of an A.P be a ,b , c, respectively . Then show that the sum is
(b+c-2a) (a+c) / 2(b-a) .
Q. If first term ,second , and last term of an A.P be a ,b , c, respectively . Then show that the sum is
[ (a+c) * (b+c - 2a) ] / 2 *(b-a) .
(n-1) = (c-a) / [b-a]
=> n = { (c-a) / [b-a] } + 1 = { (c-a) + (b-a) } / [b-a] = {b+c - 2a} / [b-a]
Therefore : n = {b+c - 2a} / [b-a] -------(1)
To determine Sum of n term (Sn) => Apply : Sn = [n / 2 ] * { a1 + an }
Sn = [n / 2 ] * { a + c } -----------(2)
From (1) : n = {b+c - 2a} / [b-a] , Substitute in (2) .
Sn = [ {n} / 2 ] * { a + c }
Sn = [ { {b+c - 2a} / [b-a] } * (1 / 2) ] * { a + c }
Sn = [ (a+c) * (b+c - 2a) ] / 2 *(b-a)
Hence Sum of n terms : Sn = [ (a+c) * (b+c - 2a) ] / 2 *(b-a)
(HENCE PROVED)
IF YOU ARE SATISFIED DO GIVE THUMBS UP.