Question

Find x so that x x+2 and x+6 are consecutive terms of a geometric progression

Answer 1

$(x + 2)(x + 6) = {b}^{2} \\ {x}^{2} + 6x + 2x + 12 = 0 \\ {x}^{2} + 8x + 12 = 0$
sole it and you will got your answer exactly right

Answer 2

Given,

x, x+2 and x+6 are in Geometric Progression.

Here,

t1 = x,

t2 = x+2,

t3 = x+6

$r = \frac{t2}{t1} = \frac{t3}{t2}$

$r = \frac{x + 2}{x} = \frac{x + 6}{x + 2}$

$( {x + 2})^{2} = x(x + 6)$

${x}^{2} + 4x + 4 = {x}^{2} + 6x$

$4x + 4 = 6x$

$4x - 6x = - 4$

$- 2x = - 4$

$x = \frac{4}{2}$

$x = 2$

Therefore the value of x is 2 such that x,x+2,x+6 are in Geometric Progression .

Additional information:

what is a Geometric progression?

The successive terms are obtained by multiplying the preceding term by a fixed number. Such type of list of numbers is said to form Geometric Progression.

1. The fixed number is called the common ration 'r' of GP.

2. The general form of GP is a,ar,ar^2,ar^3...

Example: 30,90,270,810,....