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

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(a) The sum of the first three terms of a decreasing exponential sequence (G.P) is ...

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(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}}) d x

Explanation

(a)(i)

T_{n} = a r^{n - 1}

(terms of a G.P)

T_{1} + T_{2} + T_{3} = a + a r + a r^{2} = 7 . . . (1)

a (a r) (a r^{2}) = 8 . . . . (2)

(a^{3} r^{3}) = 8 ⟹ (a r)^{3} = 2^{3}

∴ T_{2} = a r = 2

a + a r^{2} = 5 ⟹ a (1 + r^{2}) = 5

a = \frac{5}{1 + r^{2}}

(\frac{5}{1 + r^{2}}) (\frac{5 r^{2}}{1 + r^{2}}) = 4 = 2^{2}

\frac{25 r^{2}}{(1 + r^{2})^{2}} = 2^{2}

Taking square root of both sides, we have

\frac{5 r}{1 + r^{2}} = 2 ⟹ 5 r = 2 + 2 r^{2}

2 r^{2} - 5 r + 2 = 0 ⟹ 2 r^{2} - 4 r - r + 2 = 0

2 r (r - 2) - 1 (r - 2) = 0 ⟹ r = \frac{1}{2}; r = 2

Since it is a decreasing sequence, the common ratio

r = \frac{1}{2}

.
(ii)

T_{2} = 2 = a r

\frac{a}{2} = 2 ⟹ a = 4

T_{3} = a r^{2} = 4 (\frac{1}{2})^{2}

4 (\frac{1}{4}) = 1

∴ The first 3 terms of the sequence are 4, 2, 1.

(b)

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

H e i g h t (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

A r e a = \frac{1}{2} \times 1 [8.08 + 19.694]

\frac{1}{2} [27.774]

13.887 ≊ 13.89

(to 2 d.p.)

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