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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 ${b^2} = ac$ .

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 ${b^2} = ac$ .