(a) If \(f(x) = \int (4x – x^{2}) \mathrm {d} x\) and f(3) = 21, find f(x).
(b) The second, fourth and eigth terms of an Arithmetic Progression (A.P) form the first three consecutive terms of a Geometric Progression (G.P). The sum of the third and fifth terms of the A.P is 20, find the :
(i) first four terms of the A.P
(ii) sum of the first ten terms of the A.P
Explanation
(a) \(f(x) = \int (4x - x^{2}) \mathrm {d} x\)
\(f(x) = 2x^{2} - \frac{x^{3}}{3} + c\)
\(f(3) = 2(3^{2}) - \frac{3^{3}}{3} + c\)
\(21 = 18 - 9 + c\)
\(c = 21 - 9 = 12\)
\(f(x) = 2x^{2} - \frac{x^{3}}{3} + 12\)
(b) \(T_{n} = a + (n - 1) d\) (terms of an A.P)
\(T_{2} = a + d ; T_{4} = a + 3d ; T_{8} = a + 7d\)
These are consecutive terms of a G.P
\(i.e. \frac{T_{4}}{T_{2}} = \frac{T_{8}}{T_{4}}\)
\((T_{4})^{2} = T_{2} \times T_{8}\)
\((a + 3d)^{2} = (a + d)(a + 7d)\)
\(a^{2} + 6ad + 9d^{2} = a^{2} + 8ad + 7d^{2}\)
\(6ad + 9d^{2} = 8ad + 7d^{2} \implies 2ad = 2d^{2}\)
\(a = d\)
\(T_{3} + T_{5} = 20\)
\(a + 2d + a + 4d = 20\)
\(a + 2a + a + 4a = 20 \implies 8a = 20\)
\(a = d = \frac{5}{2}\)
(i) First four terms of A.P = \(\frac{5}{2}, 5, \frac{15}{2}, 10\)
(ii) \(S_{n} = \frac{n}{2} (2a + (n - 1)d)\)
\(S_{10} = \frac{10}{2} (2(\frac{5}{2}) + (10 - 1)(\frac{5}{2}))\)
= \(5(5 + \frac{45}{2})\)
= \(\frac{275}{2} = 137.5\)