# For what positive values of x, the numbers $- \dfrac{2}{7},x, - \dfrac{7}{2}$ are in GP.
Hint: Use, ${b^2} = ac$ .
We know that three numbers a, b, c is in GP if, ${b^2} = ac$ . Here $b = x,a = - \dfrac{2}{7},c = - \dfrac{7}{2}$ . Let’s substitute the values, ${x^2} = ( - \dfrac{2}{7})( - \dfrac{7}{2}) \Rightarrow {x^2} = 1 \Rightarrow x = \pm 1$ .
Hence, the numbers $- \dfrac{2}{7},x, - \dfrac{7}{2}$ to be in GP, x has to be $\pm 1$ .
Note: In this question, we need to find a second term. In general, there could be any term among these three for this kind of problem where only we need to apply ${b^2} = ac$ .