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Sum of 1 to n natural numbers is 36, then find the value of n.

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Sol.    1, 2, 3, ............., n is an A. P. of natural
          numbers where a = 1, d = 1
         Sn = 36

∴      Sn\({n\over2}[2a+(n-1)d]\)                        [Formula]

        36 = \({n\over2}[2*1+(n-1)1]\)

         36 × 2 = n [2 + n –1]
∴                             72 = n (1 + n)
∴                             72 = n + n2
                   n2 + n –72 = 0
        n2 + 9n – 8n – 72 = 0
        n(n + 9) –8(n + 9) = 0
       (n + 9) (n – 8) = 0
∴      n + 9 = 0      or    n – 8 = 0
∴            n = –9    or          n = 8
         n cannot be negative
∴      n = 8

Arithmetic Progression September 28 , 2018 0 Comments 95 views