WAEC 2010 · Paper 2 · Q13

(a)
The second, fourth and sixth terms of an Arithmetic Progression (A.P.) are $x - 1$ , $x + 1$ and $7$ respectively. Find the: (i) common difference; (ii) first term; (iii) value of $x$ .
Separate values with commas, e.g. 3, −2
(b)
A spherical bowl of radius $r\text{ cm}$ is one-quarter full when 6 litres of water is poured into it. Calculate, correct to three significant figures, its diameter. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

(a)

The $n$ th term of an A.P. is $a + (n - 1)d$ .
So $a + d = x - 1$ , $a + 3d = x + 1$ and $a + 5d = 7$ .

(i)

Take the second term from the fourth: $2d = (x + 1) - (x - 1) = 2$ , so $d = 1$ .

(ii)

Put $d = 1$ into $a + 5d = 7$ : $a = 2$ .

(iii)

Put $a = 2$ , $d = 1$ into $a + d = x - 1$ : $3 = x - 1$ , so $x = 4$ .

(b)

6 litres is $6000\text{ cm}^3$ , and this is a quarter of the volume of the sphere: $\frac14 \times \frac43 \times \frac{22}{7} \times r^3 = 6000$ .
Simplify: $\frac{22}{21} r^3 = 6000$ , so $r^3 = 6000 \times \frac{21}{22} = 5727.3$ .
Take the cube root: $r = 17.89\text{ cm}$ .
The diameter is $2r = 35.8\text{ cm}$ to 3 significant figures.

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