(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.
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)