(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...

Answer Details

(a)(i) The series \(A(1+r) + A(1+r)^2 + \dots + A(1+r)^n\) is a geometric progression with first term \(a = A(1+r)\), common ratio \(R = (1+r)\) and \(n\) terms. Its sum is

\[S_n = \frac{a(R^n - 1)}{R - 1} = \frac{A(1+r)\big[(1+r)^n - 1\big]}{(1+r) - 1} = \frac{A(1+r)\big[(1+r)^n - 1\big]}{r}.\]

(ii) With \(r = 8\% = 0.08\) and \(A = 40\), the \(k\)-th term is \(T_k = 40(1.08)^k\). The sum of the 6th to 10th terms is

\[\sum_{k=6}^{10} 40(1.08)^k = 40\big[(1.08)^6 + (1.08)^7 + (1.08)^8 + (1.08)^9 + (1.08)^{10}\big].\]

Computing the powers: \(1.586874,\ 1.713824,\ 1.850930,\ 1.999005,\ 2.158925\); their sum is \(9.309558\). Hence

\[S = 40 \times 9.309558 \approx \text{GH}\phi\,372.38.\]

The sum of the 6th to 10th terms is about GH\(\phi\)372.38.

(b) For \(y = \dfrac{1}{x} = x^{-1}\), \(\dfrac{dy}{dx} = -x^{-2} = -\dfrac{1}{x^2}\).

At \(x = 2\): gradient \(= -\dfrac{1}{4}\), and the point is \(\left(2, \tfrac12\right)\). The tangent is

\[y - \tfrac{1}{2} = -\tfrac{1}{4}(x - 2) \;\Rightarrow\; y = 1 - \tfrac{1}{4}x, \quad\text{or}\quad x + 4y = 4.\]