(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:
(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.
(a) Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).
The 8th and 4th terms are \(T_8 = a + 7d\) and \(T_4 = a + 3d\).
(i) Their difference is 20:
\[ (a + 7d) - (a + 3d) = 20 \implies 4d = 20 \implies d = 5. \]
The common difference is 5.
(ii) The 8th term is \(1\tfrac{1}{2}\) times the 4th term:
\[ a + 7d = \tfrac{3}{2}(a + 3d). \]
Substitute \(d = 5\):
\[ a + 35 = \tfrac{3}{2}(a + 15) \implies 2(a + 35) = 3(a + 15) \implies 2a + 70 = 3a + 45. \]
\[ a = 25. \]
The first term is 25.
(b) A 5% depreciation each year multiplies the value by \(0.95\) annually. After 9 years:
\[ V = 10250 \times (0.95)^{9}. \]
Now \((0.95)^{9} = 0.6302\) (to 4 d.p.), so
\[ V = 10250 \times 0.6302 = 6460.06. \]
Correct to three significant figures, the value in January 1989 is N6 460.00.