(a) The 3rd and 6th terms of a Geometric Progression (G.P) are 2 and 54 respectively. Find the : (i) common ratio ; (ii) first term ; (iii) sum of the first...
Question 1 Report
(a) The 3rd and 6th terms of a Geometric Progression (G.P) are 2 and 54 respectively. Find the : (i) common ratio ; (ii) first term ; (iii) sum of the first 10 terms, correct to the nearest whole number.
(b) The ratio of the coefficient of \(x^{4}\) to that of \(x^{3}\) in the binomial expansion of \((1 + 2x)^{n}\) is \(3 : 1\). Find the value of n.
Answer Details
(a) For a G.P., \(T_3 = ar^2 = 2\) and \(T_6 = ar^5 = 54\).
(i) Dividing: \(\dfrac{ar^5}{ar^2} = \dfrac{54}{2} \Rightarrow r^3 = 27 \Rightarrow r = 3\).
(ii) \(ar^2 = 2 \Rightarrow 9a = 2 \Rightarrow a = \dfrac{2}{9}\).
(iii) Sum of first 10 terms:
\[S_{10} = \frac{a(r^{10} - 1)}{r - 1} = \frac{\tfrac{2}{9}(3^{10} - 1)}{3 - 1} = \frac{\tfrac{2}{9}(59049 - 1)}{2} = \frac{59048}{9} \approx 6561\](b) In \((1 + 2x)^n\): coefficient of \(x^3 = \binom{n}{3}2^3\); coefficient of \(x^4 = \binom{n}{4}2^4\).
\[\frac{\text{coeff }x^4}{\text{coeff }x^3} = \frac{16\binom{n}{4}}{8\binom{n}{3}} = 2\cdot\frac{n-3}{4} = \frac{n-3}{2}\]Setting this equal to \(\dfrac{3}{1}\):
\[\frac{n - 3}{2} = 3 \;\Rightarrow\; n - 3 = 6 \;\Rightarrow\; n = 9\]Download The App On Google Playstore
Everything you need to excel in your exams
