Question

Three numbers a, b, 12 are in G.P. and a, b,9
are in A.P., then a and bare-
(A) 3,6
(B)- 3,6
(C)3, -6
(1) -3, -6​

Answer 1

Answer:

$\because \: a, \: b, \: 12 \: are \: in \: gp \\ \therefore \: \frac{b}{a} = \frac{12}{b} \implies \: {b}^{2} = 12a......(1) \\ \because \: a, \: b, \: 9 \: are \: in \: ap \\ \therefore \: b - a = 9 - b \\ b + b = a + 9 \\ 2b = 9 + a \implies b = \frac{a + 9}{2} ....(2) \\ from \: equations \: (1) \: and \: (2) \\ (\frac{a + 9}{2})^{2} = 12a \\ \frac{ {(a + 9)}^{2} }{ {2}^{2} } = 12a \\ \frac{ {a}^{2} + 18a + 81}{4} = 12a \\ {a}^{2} + 18a + 81 = 48a \\ {a}^{2} + 18a + 81 - 48a = 0 \\ {a}^{2} - 30a + 81 = 0 \\ {a}^{2} - 27a - 3a+ 81 = 0 \\ a(a - 27) - 3(a - 27) = 0 \\ (a - 27)(a - 3) = 0 \\ a - 27 = 0 \: or \: a - 3 = 0 \\ a \: = 27 \: or \: a = 3 \\ now \: when \\ a = 27\implies b = \frac{27 + 9}{2} =\frac{36}{2} = 18 \\ next \: when \\ a = 3 \: \implies b = \frac{3 + 9}{2} = \frac{12}{2} = 6 \\ so \: at \: a = 27 \: \: \: \: b = 18 \\ and \: at \: a = 3 \: \: \: \: b = 6$

so, option (A) 3, 6 is the correct answer.