Question

Find the value of x such that (i) x plus 9, x - 6 and 4 are three consecutive terms of a g.p.

Answer 1

Dear, here is the answer ...
Let a = x+9; b = x−6; 
c = 4
Since, a, b and c are in GP, then    b2 = ac
⇒(x−6)2 = 4(Let a = k+9; 
b = x−6;
 c = 4
Since, a, b and c are in GP, then   
 b2 = ac⇒(x−6)2 
= 4(x+9)⇒k2 + 36 − 12x 
= 4x+36⇒x2 − 16x
= 0⇒x(x−16)
 = 0⇒x = 0  or  x = 16+9)
⇒x2 + 36 − 12x = 4x+36
⇒x2 − 16k = 0
⇒x(x−16) =
⇒x = 0  or  x = 16

Hope it helps you (^^)^_^(^^)