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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  \({(a-c)(b+c-2a)}\over 2(b-a)\)

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Sol.    For the given A. P. a, b, ......., c
          First term = a
         Common difference d = b – a.
         Last term tn = c
∴          tn = 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)

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

          From (I) and t1 = a, tn = c

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

          Sn =   \((b + c – 2a) (a + c)\over2(b – a)\)   [From (I)]

∴       Sn =   \((a + c)(b + c – 2a)\over2(b-a)\)

∴        Hence proved

Arithmetic Progression September 28 , 2018 0 Comments 75 views