Question

Let $t_{1},t_{2},t_{3},\dots,t_{n}$ be in G.P. Then $(\frac{t_{4}}{t_{2}})^{3}$ is equal to

• A. $(\frac{t_{7}}{t_{3}})^{2}$
• B. $(\frac{t_{7}}{t_{4}})^{2}$
• C. $(\frac{t_{6}}{t_{4}})^{2}$
• D. $(\frac{t_{7}}{t_{2}})^{2}$
• E. $(\frac{t_{9}}{t_{8}})^{2}$

Hint:

Math Tip: When dealing with ratios of terms in a G.P., the first term '$a$' always cancels out. The ratio $\frac{t_m}{t_n}$ simplifies directly to $r^{m-n}$. So, $\frac{t_4}{t_2} = r^{4-2} = r^2$.

Answer 1

The Correct Option is B

Concept:
Sequences and Series - General Term of a Geometric Progression (G.P.).
The $n$-th term of a G.P. is given by $t_n = ar^{n-1}$, where $a$ is the first term and $r$ is the common ratio.
Step 1: Express the given terms using the G.P. formula.
Let the first term be $a$ and the common ratio be $r$. $$ t_2 = ar^{2-1} = ar $$ $$ t_4 = ar^{4-1} = ar^3 $$
Step 2: Evaluate the given expression.
Substitute $t_2$ and $t_4$ into the expression $(\frac{t_4}{t_2})^3$: $$ \left(\frac{ar^3}{ar}\right)^3 $$ Cancel out '$a$' and simplify the powers of '$r$': $$ (r^2)^3 = r^6 $$ The target value we are looking for in the options is $r^6$.
Step 3: Evaluate Option (A).
$$ \left(\frac{t_7}{t_3}\right)^2 = \left(\frac{ar^6}{ar^2}\right)^2 = (r^4)^2 = r^8 \quad \text{(Incorrect)} $$
Step 4: Evaluate Option (B).
$$ \left(\frac{t_7}{t_4}\right)^2 = \left(\frac{ar^6}{ar^3}\right)^2 = (r^3)^2 = r^6 \quad \text{(Correct)} $$