(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 \(\int_{1} ^{5} (x + \frac{2}{x^{2}}) \mathrm {d} x\).
(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.)
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