Question

2, x,y are in g.p and 6 is the A.M between x and y find the value of x and y

Answer 1

Answer:

$\boxed{ \pink{ \huge{x = - 6 \: and \: y = 18}}} \\ \boxed{ \pink{ \huge{x = 4 \: and \: y =8}}}$

Step-by-step explanation:

$\bold{\large{\red{\underline{ Given}}}} \longrightarrow \\ 2 ,\: x, \: y \: are \: in \: GP \: and \: \\ 6 \: is \: AM \: between \: x \: and \: y \\ \bold{ \underline{\large{\red{To \: find }}}} \longrightarrow \\ value \: of \: x \: and \: y \\ \bold{ \underline{\large{\red{Concept \: used}}}} \longrightarrow \\ if \: a \:, b \: and \: c \: are \: in \: GP \\ {b}^{2} = ac \\ if \: p \: is \: AM \: between \: a \: and \: b \\ p = \dfrac{a + b}{2} \\ \bold{ \underline{\large{\red{Solution}}}} \longrightarrow \\ 2, \: x, \: y \: are \: in \: GP \\ {x}^{2} = 2y \\ now \: 6 \: is \: AM \: of \: x \: and \: y \\ \dfrac{x + y}{2} = 6 \\ x + y = 12 \\ putting \: y \: = \frac{ {x}^{2} }{2} \: in \: it \: we \: get \\ x + \frac{ {x}^{2} }{2} = 12 \\ = > {x}^{2} + 2x = 24 \\ = > {x}^{2} + 2x - 24 = 0 \\ = > {x}^{2} + (6 - 4)x - 24 = 0 \\ = > {x}^{2} + 6x - 4x - 24 = 0 \\ = > x(x + 6) - 4(x + 6) = 0 \\ = > (x + 6)(x - 4) = 0 \\ if \: \: \: \: x + 6 = 0 \\ = > x = - 6 \\ now \\ \: \: \: \: \: \: \: {x}^{2} = 2y \\ = > {( - 6)}^{2} = 2y \\ = > 2y = 36 \\ = > y = 18 \\ if \: \: \: x - 4 = 0 \\ = > x = 4 \\ now \\ \: \: \: \: \: {x}^{2} = 2y \\ = > {(4)}^{2} = 2y \\ = > 2y = 16 \\ = > y = 8$