(a) If 3, x, y, 18 are the terms of an Arithmetic Progression (A.P), find the values of x and y.
(b)(i) The sum of the second and third terms of a grometric progression is six times the fourth term. Find the two possible values of the common ratio.
(ii) If the second term is 8 and the common ratio is positive, find the first six terms.
Explanation
(a) 3, x, y, 18.
Note : When given the consecutive terms of an A.P to be (a, b, c, d), then
(frac{a + c}{2} = b ; frac{b + d}{2} = c).
(therefore frac{3 + y}{2} = x implies 3 + y = 2x …. (1))
Also, (frac{x + 18}{2} = y implies x + 18 = 2y ….. (2))
From (2), x = 2y – 18. Putting that in (1), we have
(3 + y = 2(2y – 18) implies 3 + y = 4y – 36)
(3 + 36 = 4y – y implies 3y = 39)
(y = 13)
(x = 2y – 18)
(x = 2(13) – 18 = 26 – 18 = 8)
((x, y) = (8, 13)).
(b)(i) G.P
(T_{2} = ar ; T_{3} = ar^{2} ; T_{4} = ar^{3})
(T_{2} + T_{3} = 6T_{4})
(ar + ar^{2} = 6ar^{3})
(6ar^{3} – ar^{2} – ar = 0)
(6ar^{3} – 3ar^{2} + 2ar^{2} – ar = 0)
(3ar^{2}(2r – 1) + ar(2r – 1) = 0)
((3ar^{2} + ar)(2r – 1) = 0)
(ar(3r + 1)(2r – 1) = 0)
(implies 3r + 1 = 0 ; 2r – 1 = 0)
(r = frac{-1}{3} ; r = frac{1}{2})
(ii) Since r is positive, then (r = frac{1}{2}).
(T_{2} = ar = 8 implies frac{a}{2} = 8)
(a = 16)
(therefore text{The first 6 terms} = 16, 8, 4, 2, 1, frac{1}{2})