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

Step 1: Understanding the Question: 
We are given the 3rd and 6th terms of a Geometric Progression (GP). We need to find the sum of the first 5 terms (S\(_{5}\)). 
Step 2: Key Formula or Approach: 
The formula for the n\(^{th}\) term of a GP is a\(_{n}\) = ar\(^{n-1}\), where 'a' is the first term and 'r' is the common ratio. 
The formula for the sum of the first n terms of a GP is S\(_{n}\) = \(\frac{a(r^n - 1)}{r - 1}\). 
Step 3: Detailed Explanation: 
We are given: 
3rd term, a\(_{3}\) = ar\(^{3-1}\) = ar\(^{2}\) = 12   --- (1) 
6th term, a\(_{6}\) = ar\(^{6-1}\) = ar\(^{5}\) = 96   --- (2) 
To find the common ratio 'r', we divide equation (2) by equation (1): 
\[ \frac{ar^5}{ar^2} = \frac{96}{12} \] \[ r^3 = 8 \] \[ r = \sqrt[3]{8} = 2 \] Now, substitute the value of r = 2 back into equation (1) to find the first term 'a': 
\[ a(2)^2 = 12 \] \[ 4a = 12 \] \[ a = 3 \] Now we have the first term (a = 3) and the common ratio (r = 2). We can find the sum of the first 5 terms, S\(_{5}\): 
\[ S_5 = \frac{a(r^5 - 1)}{r - 1} \] \[ S_5 = \frac{3(2^5 - 1)}{2 - 1} \] \[ S_5 = \frac{3(32 - 1)}{1} \] \[ S_5 = 3(31) = 93 \] Step 4: Final Answer: 
The sum of the first 5 terms of the Geometric Progression is 93. 
 

Answer 2

Concept:

• The ratio of two GP terms removes the first term.
• Once the common ratio is known, the required terms can be written directly.

Step 1: Find the common ratio.
$\dfrac{T_6}{T_3}=r^3=\dfrac{96}{12}=8$, so $r=2$.

Step 2: Move backward from the third term.
Since $T_3=12$, $T_2=12/2=6$ and $T_1=6/2=3$.

Step 3: Write and add the first five terms.
The terms are $3,6,12,24,48$. Their sum is $3+6+12+24+48=93$.

Final Answer: $93$, option A