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WAEC 2013 · Paper 2 · Q11

  1. (a)

    The sum of the first three terms of a decreasing exponential sequence (G.P.) is equal to 7 and the product of these three terms is equal to 8. Find the: (i) common ratio; (ii) first three terms of the sequence.

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

  2. (b)

    Using the trapezium rule with ordinates at x=1,2,3,4x = 1, 2, 3, 4 and 55, calculate, correct to two decimal places, the value of ∫15(x+2x2)dx\displaystyle\int_1^5 \left(x + \frac{2}{x^2}\right)dx.

Worked solution (try it first)

(a)(i)

  1. Write the three terms as ar\dfrac ar, aa and arar.
  2. Their product is a3=8a^3 = 8, so a=2a = 2.
  3. Their sum: 2r+2+2r=7\dfrac2r + 2 + 2r = 7.
  4. Multiply by rr: 2+2r+2r2=7r2 + 2r + 2r^2 = 7r, so 2r2−5r+2=02r^2 - 5r + 2 = 0.
  5. Factorise: (2r−1)(r−2)=0(2r - 1)(r - 2) = 0, so r=12r = \frac12 or r=2r = 2.
  6. The sequence is decreasing, so r=12r = \frac12.

(ii)

  1. The terms are 212=4\dfrac{2}{\frac12} = 4, then 22, then 11.

(b)

  1. h=1h = 1.
  2. y=x+2x2y = x + \dfrac{2}{x^2} gives 3, 2.5, 3.2222, 4.125, 5.083,\ 2.5,\ 3.2222,\ 4.125,\ 5.08 at x=1,…,5x = 1, \ldots, 5.
  3. 12[(3+5.08)+2(2.5+3.2222+4.125)]=12[8.08+19.6944]\frac12[(3 + 5.08) + 2(2.5 + 3.2222 + 4.125)] = \frac12[8.08 + 19.6944]
    =13.8872= 13.8872, about 13.8913.89.

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