(a) In an A.P, the difference between the 8th and 4th terms is 20 and the 8th term is \(1\frac{1}{2}\) 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.
(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.
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Correct Answer: Option n
Explanation:
(a) \(T_{n} = a + (n - 1) d\) (terms of an A.P)
Given: \(T_{8} - T_{4} = 20\)
\(T_{8} = 1\frac{1}{2} \times T_{4}\)
\(\therefore a + 7d - (a + 3d) = 20\)
\(4d = 20 \implies d = 5\)
(ii) Put d = 5 in the equation
\(a + 7d = 1\frac{1}{2} \times (a + 3d)\)
\(a + 7d = 1.5a + 4.5d\)
\(7d - 4.5d = 1.5a - a \implies 2.5d = 0.5a\)
\(a = \frac{2.5 \times 5}{0.5} = 25\)
(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)
(a) \(T_{n} = a + (n - 1) d\) (terms of an A.P)
Given: \(T_{8} - T_{4} = 20\)
\(T_{8} = 1\frac{1}{2} \times T_{4}\)
\(\therefore a + 7d - (a + 3d) = 20\)
\(4d = 20 \implies d = 5\)
(ii) Put d = 5 in the equation
\(a + 7d = 1\frac{1}{2} \times (a + 3d)\)
\(a + 7d = 1.5a + 4.5d\)
\(7d - 4.5d = 1.5a - a \implies 2.5d = 0.5a\)
\(a = \frac{2.5 \times 5}{0.5} = 25\)
(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)