Stepwise
Start practisingPractise

WAEC 2009 · Paper 2 · Q12

  1. (a)

    Three positive numbers are in arithmetic progression (A.P.). The sum of the squares of the three numbers is 155, while the sum of the numbers is 21. If the common difference is positive, find the numbers.

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

  2. (b)

    If the total surface area of a sphere is 154 cm2154\text{ cm}^2, find its radius. [Take π=227\pi = \frac{22}{7}]

Worked solution (try it first)

(a)

  1. Write the numbers as a−da - d, aa, a+da + d.
  2. Their sum is 3a=213a = 21, so a=7a = 7.
  3. Sum of squares: (7−d)2+49+(7+d)2=155(7 - d)^2 + 49 + (7 + d)^2 = 155.
  4. Expand: 49−14d+d2+49+49+14d+d2=147+2d249 - 14d + d^2 + 49 + 49 + 14d + d^2 = 147 + 2d^2
    =155= 155.
  5. So 2d2=82d^2 = 8, d2=4d^2 = 4 and d=2d = 2 (positive).
  6. The numbers are 5, 7 and 9 (check: 25+49+81=15525 + 49 + 81 = 155).

(b)

  1. Surface area of a sphere =4πr2= 4\pi r^2: 4×227×r2=1544 \times \frac{22}{7} \times r^2 = 154.
  2. r2=154×788=12.25r^2 = \frac{154 \times 7}{88} = 12.25, so r=3.5r = 3.5 cm.

Report a problem with this question