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WAEC 2008 · Paper 2 · Q13

  1. (a)

    If 3,x,y,183, x, y, 18 are in arithmetic progression (A.P.), find the values of xx and yy.

    Separate values with commas, e.g. 3, −2

  2. (b)

    (i) The sum of the second and third terms of a geometric progression is six times the fourth term. Find the two possible values of the common ratio. (ii) If the second term is 88 and the common ratio is positive, find the first six terms.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. 1818 is the fourth term, so 3+3d=183 + 3d = 18 and the common difference is d=5d = 5.
  2. Then x=3+5=8x = 3 + 5 = 8 and y=8+5=13y = 8 + 5 = 13.

(b)(i)

  1. With first term aa and ratio rr: ar+ar2=6ar3ar + ar^2 = 6ar^3.
  2. Divide by arar (both are non-zero): 1+r=6r21 + r = 6r^2, so 6r2−r−1=06r^2 - r - 1 = 0.
  3. Factorise: (3r+1)(2r−1)=0(3r + 1)(2r - 1) = 0, so r=12r = \frac12 or r=−13r = -\frac13.

(ii)

  1. The positive ratio is r=12r = \frac12.
  2. The second term is ar=8ar = 8, so a=16a = 16.
  3. Halve each time: the first six terms are 16,8,4,2,1,1216, 8, 4, 2, 1, \frac12.

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