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WAEC 2011 · Paper 2 · Q4

  1. (a)

    Find the third term of the exponential sequence (GP) (2−1),(3−22),…(\sqrt2 - 1), (3 - 2\sqrt2), \ldots

Worked solution (try it first)
  1. The common ratio is the second term divided by the first: r=3−222−1r = \dfrac{3 - 2\sqrt2}{\sqrt2 - 1}.
  2. Notice that (2−1)2=2−22+1(\sqrt2 - 1)^2 = 2 - 2\sqrt2 + 1
    =3−22= 3 - 2\sqrt2.
  3. So the second term is the first term squared, and r=2−1r = \sqrt2 - 1.
  4. The third term is the second term times rr: T3=(3−22)(2−1)T_3 = (3 - 2\sqrt2)(\sqrt2 - 1).
  5. Expand: 32−3−2×2+22=32−3−4+223\sqrt2 - 3 - 2 \times 2 + 2\sqrt2 = 3\sqrt2 - 3 - 4 + 2\sqrt2.
  6. Collect the terms: T3=52−7T_3 = 5\sqrt2 - 7.

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