Given that the first and fourth terms of a G.P are 6 and 162 respectively, find the sum of the first three terms of the progression.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option D
Explanation:
To find the sum of the first three terms of a geometric progression (G.P.) where the first term (\(a\)) is 6 and the fourth term is 162, follow these steps:
1. Identify the given values and the common ratio (\(r\)):
- The first term \(a = 6\).
- The fourth term is given by \(a \cdot r^3 = 162\).
2. Set up the equation for the fourth term:
\[
6 \cdot r^3 = 162
\]
Solve for \(r^3\):
\[
r^3 = \frac{162}{6} = 27
\]
Find \(r\):
\[
r = \sqrt[3]{27} = 3
\]
3. Calculate the first three terms of the G.P.:
- The first term is \(a = 6\).
- The second term is \(a \cdot r = 6 \cdot 3 = 18\).
- The third term is \(a \cdot r^2 = 6 \cdot 3^2 = 6 \cdot 9 = 54\).
4. Find the sum of the first three terms:
\[
\text{Sum} = 6 + 18 + 54 = 78
\]
Thus, the sum of the first three terms of the geometric progression is 78, which corresponds to option D.
To find the sum of the first three terms of a geometric progression (G.P.) where the first term (\(a\)) is 6 and the fourth term is 162, follow these steps:
1. Identify the given values and the common ratio (\(r\)):
- The first term \(a = 6\).
- The fourth term is given by \(a \cdot r^3 = 162\).
2. Set up the equation for the fourth term:
\[
6 \cdot r^3 = 162
\]
Solve for \(r^3\):
\[
r^3 = \frac{162}{6} = 27
\]
Find \(r\):
\[
r = \sqrt[3]{27} = 3
\]
3. Calculate the first three terms of the G.P.:
- The first term is \(a = 6\).
- The second term is \(a \cdot r = 6 \cdot 3 = 18\).
- The third term is \(a \cdot r^2 = 6 \cdot 3^2 = 6 \cdot 9 = 54\).
4. Find the sum of the first three terms:
\[
\text{Sum} = 6 + 18 + 54 = 78
\]
Thus, the sum of the first three terms of the geometric progression is 78, which corresponds to option D.