If the 4^{th}, 7^{th} and 10^{th} terms of a | vedclass

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(C) Let the first term of the $G.P.$ be $A$ and the common ratio be $r$.We know that the $n^{th}$ term of a $G.P.$ is given by $T_n = Ar^{n-1}$.Given

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$b^2 = ac$

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(C) Let the first term of the $G.P.$ be $A$ and the common ratio be $r$.We know that the $n^{th}$ term of a $G.P.$ is given by $T_n = Ar^{n-1}$.Given that $T_4 = a$,$T_7 = b$,and $T_{10} = c$,we have:$a = Ar^3$$b = Ar^6$$c = Ar^9$Now,consider the product $ac$:$ac = (Ar^3)(Ar^9) = A^2r^{12}$Also,consider $b^2$:$b^2 = (Ar^6)^2 = A^2r^{12}$Since $b^2 = A^2r^{12}$ and $ac = A^2r^{12}$,it follows that $b^2 = ac$.Alternatively,if the indices $p, q, r$ are in $A.P.$,then the $p^{th}, q^{th}, r^{th}$ terms of a $G.P.$ are in $G.P.$. Here,$4, 7, 10$ are in $A.P.$ with a common difference of $3$,therefore $a, b, c$ are in $G.P.$,which implies $b^2 = ac$.

If the $4^{th}, 7^{th}$ and $10^{th}$ terms of a $G.P.$ are $a, b, c$ respectively,then the relation between $a, b, c$ is

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