Question

In a Geometric Progression, 3\(^{rd}\) term is 12, and 6\(^{th}\) term is 96. Find sum of first 5 terms.

• A. 93
• B. 186
• C. 248
• D. 124

Hint:

Divide the sixth-term equation by the third-term equation to obtain $r^3$. Then either list the first five terms or use the finite GP sum formula.

Answer 1

The Correct Option is A

Concept:

• Represent GP terms as $ar^{n-1}$.
• Use the finite GP sum after solving the two term equations.

Step 1: Form the equations.
$ar^2=12$ and $ar^5=96$.

Step 2: Solve for $r$ and $a$.
Dividing gives $r^3=8$, so $r=2$. Then $4a=12$, hence $a=3$.

Step 3: Apply the sum formula.
$S_5=a\dfrac{r^5-1}{r-1}=3\dfrac{32-1}{1}=93$.

Final Answer: $93$