(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)