Question

If the $3^{\text{rd}}$, $7^{\text{th}}$, $11^{\text{th}}$ terms of a geometric progression are $a$, $b$, $c$ respectively, then $(ac)^{4} =$

• A. $b^{8}$
• B. $b^{6}$
• C. $b^{4}$
• D. $b^{2}$
• E. $b^{12}$

Hint:

Logic Tip: In a G.P., if the indices of three terms form an Arithmetic Progression (like 3, 7, 11 where the difference is 4), the terms themselves form a G.P., meaning the middle term squared equals the product of the extremes: $b^2 = ac$.

Answer 1

The Correct Option is A

Concept:
For a Geometric Progression (G.P.) with first term $A$ and common ratio $R$, the $n$-th term is given by $t_n = A \cdot R^{n-1}$.
Step 1: Express the given terms using the G.P. formula.
Given: $t_3 = a \implies a = A \cdot R^2$ $t_7 = b \implies b = A \cdot R^6$ $t_{11} = c \implies c = A \cdot R^{10}$
Step 2: Find the product of a and c.
$$ac = (A \cdot R^2)(A \cdot R^{10})$$ $$ac = A^2 \cdot R^{12}$$
Step 3: Express ac in terms of b.
Notice that $b^2 = (A \cdot R^6)^2 = A^2 \cdot R^{12}$. Therefore, $ac = b^2$.
Step 4: Calculate the final expression.
We need the value of $(ac)^4$. Substituting $b^2$ for $ac$: $$(ac)^4 = (b^2)^4 = b^8$$