(a) In an A.P, the difference between the 8th and 4th terms is 20 and the 8th term is (1frac{1}{2}) times the 4th term. What is the:
(i) common difference ; (ii) first term of the sequence?
(b) The value of a machine depreciates each year by 5% of its value at the beginning of that year. If its value when new on 1st January 1980 was N10,250.00, what was its value in January 1989 when it was 9 years old? Give your answer correct to three significant figures.
Explanation
(a) (T_{n} = a + (n – 1) d) (terms of an A.P)
Given: (T_{8} – T_{4} = 20)
(T_{8} = 1frac{1}{2} times T_{4})
(therefore a + 7d – (a + 3d) = 20)
(4d = 20 implies d = 5)
(ii) Put d = 5 in the equation
(a + 7d = 1frac{1}{2} times (a + 3d))
(a + 7d = 1.5a + 4.5d)
(7d – 4.5d = 1.5a – a implies 2.5d = 0.5a)
(a = frac{2.5 times 5}{0.5} = 25)
(b) (A = P(1 – frac{r}{100})^{n})
where P = N10,250.00
r = 5%
n = 9
(therefore A = (10,250) (1 – frac{5}{100})^{9})
= (10,250 times (0.95)^{9} = 10,250 times 0.6302)
= (N6,460.056 approxeq N6,460) (to 3 sig. figs)