(a) If 3, x, y, 18 are the terms of an Arithmetic Progression (A.P), find
(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 \(r = \frac{1}{2}\).
\(T_{2} = ar = 8 \implies \frac{a}{2} = 8\)
\(a = 16\)
\(\therefore \text{The first 6 terms} = 16, 8, 4, 2, 1, \frac{1}{2}\)