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If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is \({(a-c)(b+c-2a)}\over 2(b-a)\)

Print or Save**Sol**. For the given A. P. a, b, ......., c

First term = a

Common difference d = b – a.

Last term tn = c

∴ t_{n} = a + (n –1)d

c = a + (n –1) (b – a)

c – a = (n –1) (b – a)

n – 1 = \((c – a)\over(b-a)\)

n = \((c – a)\over(b-a)\)+1

n = \(c – a + b – a\over(b-a)\) [Taking LCM)

n = \(b + c – 2a\over{(b-a)}\) ...(I)

S_{n} = \({n\over2}[t1+tn]\) [Formula]

From (I) and t_{1} = a, t_{n} = c

S_{n }= \({{b + c – 2a}\over2(b – a)}[a+c]\)

S_{n} = \((b + c – 2a) (a + c)\over2(b – a)\) [From (I)]

∴ S_{n} = \((a + c)(b + c – 2a)\over2(b-a)\)

∴ Hence proved