WAEC 2010 · Paper 2 · Q2

  1. (a)

    If the sum of the first nn terms of the series 4+7+10+…4 + 7 + 10 + \dots is 209209, find nn.

Worked solution (try it first)
  1. The series is an A.P. with a=4a = 4 and d=3d = 3.
  2. Use Sn=n2[2a+(n−1)d]S_n = \dfrac n2[2a + (n - 1)d]: n2[8+3(n−1)]=209\dfrac n2[8 + 3(n - 1)] = 209.
  3. Simplify the bracket: n2(3n+5)=209\dfrac n2(3n + 5) = 209.
  4. Multiply by 2 and rearrange: 3n2+5n−418=03n^2 + 5n - 418 = 0.
  5. Factorise: (3n+38)(n−11)=0(3n + 38)(n - 11) = 0.
  6. nn must be a positive whole number, so n=11n = 11.

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