(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.)