(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) c...
Question 1 Report
(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 \(\int_{1} ^{5} (x + \frac{2}{x^{2}}) \mathrm {d} x\).
Answer Details
(a) Let the three terms be \( \dfrac{a}{r},\ a,\ ar \).
Product: \( \dfrac{a}{r}\cdot a \cdot ar = a^3 = 8 \Rightarrow a = 2. \)
Sum: \( \dfrac{2}{r} + 2 + 2r = 7 \Rightarrow \dfrac{2}{r} + 2r = 5 \Rightarrow 2r^2 - 5r + 2 = 0. \)
\[ (2r - 1)(r - 2) = 0 \Rightarrow r = \tfrac12 \text{ or } r = 2. \]
(i) The sequence is decreasing, so \( r = \dfrac12 \).
(ii) The three terms are \( \dfrac{a}{r} = 4,\ a = 2,\ ar = 1 \), i.e. \( 4,\ 2,\ 1 \).
(b) With \( f(x) = x + \dfrac{2}{x^2} \) and ordinates at \( x = 1,2,3,4,5 \) (\(h = 1\)):
\( f(1)=3,\ f(2)=2.5,\ f(3)=3.2222,\ f(4)=4.125,\ f(5)=5.08. \)
Trapezium rule: \[ \int_1^5 f\,dx \approx \frac{h}{2}\big[f(1)+f(5) + 2(f(2)+f(3)+f(4))\big]. \]
\[ = \frac12\big[3 + 5.08 + 2(2.5 + 3.2222 + 4.125)\big] = \frac12\big[8.08 + 19.6944\big] = \frac12(27.7744) \approx 13.89. \]
Download The App On Google Playstore
Everything you need to excel in your exams
