(a)(i) Find the sum of the series \(A(1 + r) + A(1 + r)^{2} +
(a)(i) Find the sum of the series \(A(1 + r) + A(1 + r)^{2} + ... + A(1 + r)^{n}\).
(ii) Given that r = 8% and A = GH 40.00, find the sum of the 6th to 10th terms of the series in (i).
(b) Find the equation of the tangent to the curve \(y = \frac{1}{x}\) at the point on the curve when x = 2.
Explanation
(a)(i) \(A(1 + r) + A(1 + r)^{2} + ... + A(1 + r)^{n}\)
\(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\) (sum of terms in a G.P).
\(a = A(1 + r) ; r = (1 + r)\)
\(S_{n} = \(\frac{(A(1 + r))[(1 + r)^{n} - 1]}{(1 + r) - 1}\)
= \(\frac{(A(1 + r))[(1 + r)^{n} - 1]}{r}\)
(ii) Sum of the 6th to 10th term = \(S_{10} - S_{5}\)
\(a = 40(1 + 0.08) = 43.2 ; r = 0.08\)
\(S_{10} = \frac{43.2[(1 + 0.08)^{10} - 1]}{0.08} = 540[(1.08)^{10} - 1]\)
\(S_{5} = \frac{43.2[(1.08)^{5} - 1]}{0.08} = 540[(1.08)^{5} - 1]\)
\(S_{10} - S_{5} = 540[(1.08)^{10} - (1.08)^{5}]\)
= 371.8
(b) \(y = \frac{1}{x} = x^{-1}\)
\(\frac{\mathrm d y}{\mathrm d x} = -x^{-2}\)
When \(x = 2; y = \frac{1}{2}\), a point on the curve.
Gradient of tangent = \(-2^{-2} = -\frac{1}{4}\)
Equation : \(\frac{y - \frac{1}{2}}{x - 2} = -\frac{1}{4}\)
\(4y - 2 = 2 - x \implies x + 4y - 4 = 0\)