If x, x+2, x+3 are consecutive terms of a GP, then find x.

Hint: Here we go through by the property of geometric progression as we know in a G.P the ratio between any two consecutive terms remains the same. By applying this statement we will solve the question.

Complete step-by-step answer:
Here in the question it is given that x, x+2, x+3 are consecutive terms of a G.P.

Now we have to find the value of x.

As we know in a G.P., the ratio between any two consecutive terms remains same and for the given three consecutive terms we can write it as,

$

   \Rightarrow \dfrac{{x + 3}}{{x + 2}} = \dfrac{{x + 2}}{x} \\

   \Rightarrow x(x + 3) = {(x + 2)^2} \\

 $

As we know ${(a + b)^2} = {a^2} + {b^2} + 2ab$ similarly we expand ${(x + 2)^2}$

$

   \Rightarrow {x^2} + 3x = {x^2} + 4 + 4x \\

   \Rightarrow 3x - 4x = 4 \\

   \Rightarrow - x = 4 \\

  \therefore x = 4 \\

 $

Hence the value of x is 4.

Note: Whenever we face such a type of question the key concept for solving the question is to go through the basic properties of the given series. Here in this question we go through with basic properties of GP. We can also solve this question with the help of geometric mean as we know if three numbers a, b and c are in GP then we write ${b^2} = ac$ by the properties of geometric mean.