(a) The sum of the first three terms of a decreasing exponential sequence (G.P) is equal to 7 and the product of these three is equal to 8. Find the :
(i) common ratio ; (ii) first three terms of the sequence.
(b) Using the trapezium rule with the ordinates at x = 1, 2, 3, 4 and 5, calculate, correct to two decimal places, the value of (int_{1} ^{5} (x + frac{2}{x^{2}}) mathrm {d} x).
Explanation
(a)(i) (T_{n} = ar^{n – 1}) (terms of a G.P)
(T_{1} + T_{2} + T_{3} = a + ar + ar^{2} = 7 … (1))
(a(ar)(ar^{2}) = 8 …. (2))
((a^{3} r^{3}) = 8 implies (ar)^{3} = 2^{3})
(therefore T_{2} = ar = 2)
(a + ar^{2} = 5 implies a(1 + r^{2}) = 5)
(a = frac{5}{1 + r^{2}})
((frac{5}{1 + r^{2}})(frac{5r^{2}}{1 + r^{2}}) = 4 = 2^{2})
(frac{25r^{2}}{(1 + r^{2})^{2}} = 2^{2})
Taking square root of both sides, we have
(frac{5r}{1 + r^{2}} = 2 implies 5r = 2 + 2r^{2})
(2r^{2} – 5r + 2 = 0 implies 2r^{2} – 4r – r + 2 = 0)
(2r(r – 2) – 1(r – 2) = 0 implies r = frac{1}{2} ; r = 2)
Since it is a decreasing sequence, the common ratio (r = frac{1}{2}).
(ii) (T_{2} = 2 = ar)
(frac{a}{2} = 2 implies a = 4)
(T_{3} = ar^{2} = 4(frac{1}{2})^{2})
= (4(frac{1}{4}) = 1)
(therefore text{The first 3 terms of the sequence are } 4, 2, 1.)
(b) (int_{1} ^{5} (x + frac{2}{x^{2}}) mathrm {d} x)
| x | 1 | 2 | 3 | 4 | 5 |
| (x^{2}) | 1.0 | 4 | 9 | 16 | 25 |
| (frac{2}{x^{2}}) | 2.0 | 0.5 | 0.222 | 0.125 | 0.08 |
| (x + frac{2}{x^{2}}) | 3.0 | 2.5 | 3.222 | 4.125 | 5.08 |
(Height (h) = 1 ; y_{1} = 3.0 ; y_{5} = 5.08)
(y_{2} = 2.5 ; y_{3} = 3.222 ; y_{4} = 4.125)
(y_{1} + y_{5} = 3.0 + 5.08 = 8.08)
(y_{2} + y_{3} + y_{4} = 2.5 + 3.222 + 4.125 = 9.847)
(2(y_{2} + y_{3} + y_{4}) = 2(9.847) = 19.694)
(Area = frac{1}{2} times 1[8.08 + 19.694])
= (frac{1}{2} [27.774])
= (13.887 approxeq 13.89) (to 2 d.p.)