Concept:1). The general form of terms of a GP is a, ar, ar2, ar3, and so on. Here, a is the first term and r is the common ratio. The nth term of a GP is Tn = arn-1 Common ratio = r = Tn/ Tn-12). If a, b and c are three quantities in GP, then and b is the geometric mean of a and c. This can be written as b2 = ac or b =?acCalculation:Given:a, b, c are in GP where a > 0, b > 0, c > 0Statement I: a2, b2, c2 are in GPAs b is G.M. of a and c, b2 = acSquaring on both sides ? b4 = a2c2? (b2)2 = (a2)(c2) ? b2 is G.M. of a2 and c2a2, b2, and c2 are in GPStatement II: \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) are in GPLet the common ratio be r.? b = ar? c = br = ar2Now, \(\displaystyle \frac{\frac{1}{b}}{\frac{1}{a}}=\frac{a}{b}=\frac{a}{ar}=\frac{1}{r}\)? \(\displaystyle \frac{\frac{1}{c}}{\frac{1}{b}}=\frac{b}{c}=\frac{ar}{ar^2}=\frac{1}{r}\)1/a,1/b,1/c are in G.P.Statement III: \(? {a}, ?{b}, ?{c} \) are in GPAs b is G.M. of a and c, ?\({b} = ?{ac}\)?Taking square roots on both sides ? \( ?{b} = ?{?{a} ?{c}} \)? ?b is GM of ?a and ?c \(? {a}, ?{b}, \ and\ ?{c} \) are in GP? Statement I, II, and III are correct.