(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.
(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.
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Correct Answer: Option n
Explanation:
(a)(i) \(T_{3} = 2 ; ar^{2} = 2 ... (i)\)
\(T_{6} = 54 ; ar^{5} = 54 ... (ii)\)
\(\frac{T_{6}}{T_{3}} = \frac{ar^{5}}{ar^{2}} = \frac{54}{2}\)
\(r^{3} = 27 \implies r = 3 \)
(ii) First term = a
\(T_{3} = ar^{2} = 2\)
\(a = \frac{2}{3^{2}} = \frac{2}{9}\)
(iii) \(S_{10} = \frac{a(r^{n} - 1)}{r - 1}\)
= \(\frac{2(3^{10} - 1)}{9(3 - 1)}\)
= \(\frac{3^{10} - 1}{9}\)
= \(\frac{59049 - 1}{9} \approxeq 6561\) (nearest whole number).
(b) \((1 + 2x)^{n} = 1 + ^{n}C_{1} (1)^{n - 1} (2x) + ^{n}C_{2} (1)^{n - 2} (2x)^{2} + ^{n}C_{3} (1)^{n - 3} (2x)^{3} + ^{n}C_{4} (1)^{n - 4} (2x)^{4}+ ...\)
\(\frac{\text{coefficient of} x^{4}}{\text{coefficient of } x^{3}} = \frac{3}{1}\)
\(\frac{n!}{4! (n - 4)!} \times 2^{4} \times \frac{3! (n - 3)!}{n!} = \frac{3}{1}\)
\(\frac{16 \times 3! (n - 3)(n - 4)!}{4 \times 3! (n - 4)! \times 8} = \frac{3}{1}\)
\(\frac{2(n - 3)}{4} = \frac{3}{1} \implies 2n - 6 = 12\)
\(2n = 18 \implies n = 9\)
(a)(i) \(T_{3} = 2 ; ar^{2} = 2 ... (i)\)
\(T_{6} = 54 ; ar^{5} = 54 ... (ii)\)
\(\frac{T_{6}}{T_{3}} = \frac{ar^{5}}{ar^{2}} = \frac{54}{2}\)
\(r^{3} = 27 \implies r = 3 \)
(ii) First term = a
\(T_{3} = ar^{2} = 2\)
\(a = \frac{2}{3^{2}} = \frac{2}{9}\)
(iii) \(S_{10} = \frac{a(r^{n} - 1)}{r - 1}\)
= \(\frac{2(3^{10} - 1)}{9(3 - 1)}\)
= \(\frac{3^{10} - 1}{9}\)
= \(\frac{59049 - 1}{9} \approxeq 6561\) (nearest whole number).
(b) \((1 + 2x)^{n} = 1 + ^{n}C_{1} (1)^{n - 1} (2x) + ^{n}C_{2} (1)^{n - 2} (2x)^{2} + ^{n}C_{3} (1)^{n - 3} (2x)^{3} + ^{n}C_{4} (1)^{n - 4} (2x)^{4}+ ...\)
\(\frac{\text{coefficient of} x^{4}}{\text{coefficient of } x^{3}} = \frac{3}{1}\)
\(\frac{n!}{4! (n - 4)!} \times 2^{4} \times \frac{3! (n - 3)!}{n!} = \frac{3}{1}\)
\(\frac{16 \times 3! (n - 3)(n - 4)!}{4 \times 3! (n - 4)! \times 8} = \frac{3}{1}\)
\(\frac{2(n - 3)}{4} = \frac{3}{1} \implies 2n - 6 = 12\)
\(2n = 18 \implies n = 9\)