(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)