WAEC 2009 · Paper 2 · Q13

Sequences & series (AP, GP)

(a)
How many numbers between 75 and 500 are divisible by 7?
(b)
The 8th term of an arithmetic progression (A.P.) is 5 times the third term, while the 7th term is 9 greater than the 4th term. Write down the first five terms of the A.P.
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

The multiples of 7 between 75 and 500 form an A.P.: the first is $77$ ( $7 \times 11$ ) and the last is $497$ ( $7 \times 71$ ), with $d = 7$ .
Use $T_n = a + (n - 1)d$ : $497 = 77 + (n - 1) \times 7$ .
So $(n - 1) \times 7 = 420$ , $n - 1 = 60$ and $n = 61$ .
There are 61 such numbers.

(b)

"7th term is 9 greater than the 4th": $a + 6d = a + 3d + 9$ , so $3d = 9$ and $d = 3$ .
"8th term is 5 times the 3rd": $a + 7d = 5(a + 2d)$ , so $a + 7d = 5a + 10d$ and $4a = -3d$ .
Put in $d = 3$ : $4a = -9$ , so $a = -\frac94$ .
Add 3 each time: the first five terms are $-\frac94, \frac34, \frac{15}{4}, \frac{27}{4}, \frac{39}{4}$ .

Report a problem with this question