Question

If the second and fifth terms of a G.P. are 24 and 3 respectively, then the sum of first six terms is:

• A. \( 181 \)
• B. \( \frac{181}{2} \)
• C. \( 189 \)
• D. \( \frac{189}{2} \)
• E. \( 191 \)

Hint:

When $r < 1$, the sum of terms will always be less than $a / (1-r)$. Here, the infinite sum would be $48 / (1/2) = 96$. This helps you quickly rule out options (A), (C), and (E).

Answer 1

The Correct Option is D

Concept: For a Geometric Progression (G.P.):
• \( n^{th} \) term \( a_n = ar^{n-1} \)
• Sum of \( n \) terms \( S_n = \frac{a(1 - r^n)}{1 - r} \)

Step 1: Find the common ratio \( r \) and first term \( a \).
Given \( a_2 = ar = 24 \) and \( a_5 = ar^4 = 3 \). Divide \( a_5 \) by \( a_2 \): \[ \frac{ar^4}{ar} = \frac{3}{24} \quad \Rightarrow \quad r^3 = \frac{1}{8} \quad \Rightarrow \quad r = \frac{1}{2} \] Substitute \( r = 1/2 \) into \( ar = 24 \): \[ a(1/2) = 24 \quad \Rightarrow \quad a = 48 \]

Step 2: Calculate the sum of the first six terms (\( S_6 \)).
\[ S_6 = \frac{48(1 - (1/2)^6)}{1 - 1/2} = \frac{48(1 - 1/64)}{1/2} \] \[ S_6 = 96 \left( \frac{63}{64} \right) \]

Step 3: Simplify the fraction.
Divide both 96 and 64 by 32: \[ S_6 = 3 \times \frac{63}{2} = \frac{189}{2} \]