(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 : \(\frac{y - \frac{1}{2}}{x - 2} = -\frac{1}{4}\) \(4y - 2 = 2 - x \implies x + 4y - 4 = 0\)