Question

The $3^{\text{rd}}$ and $6^{\text{th}}$ terms of a G.P. are, respectively 108 and -32. Then the first term is

• A. -81
• B. -243
• C. 243
• D. 81
• E. 27

Hint:

Logic Tip: In a G.P., dividing an advanced term by an earlier term (like $a_6 / a_3$) will always yield $r^k$, where $k$ is the difference in their positions ($6 - 3 = 3$). This immediately isolates the ratio.

Answer 1

The Correct Option is C

Concept:
For a Geometric Progression (G.P.), the $n$-th term is given by the formula: $$a_n = a \cdot r^{n-1}$$ where $a$ is the first term and $r$ is the common ratio.
Step 1: Set up equations based on the given terms.
[cite_start]We are given $a_3 = 108$ and $a_6 = -32$[cite: 109]. Using the general formula: Eq. 1: $a \cdot r^2 = 108$ Eq. 2: $a \cdot r^5 = -32$
Step 2: Solve for the common ratio (r).
Divide Eq. 2 by Eq. 1 to eliminate $a$: $$\frac{a \cdot r^5}{a \cdot r^2} = \frac{-32}{108}$$ $$r^3 = -\frac{32}{108}$$ Simplify the fraction by dividing the numerator and denominator by 4: $$r^3 = -\frac{8}{27}$$ Taking the cube root of both sides gives: $$r = -\frac{2}{3}$$
Step 3: Substitute r back into Eq. 1 to find a.
$$a \left(-\frac{2}{3}\right)^2 = 108$$ $$a \left(\frac{4}{9}\right) = 108$$ Multiply both sides by $\frac{9}{4}$: $$a = 108 \cdot \frac{9}{4}$$ $$a = 27 \cdot 9$$ $$a = 243$$ [cite_start]This matches Option C[cite: 114, 118].