Question

The sum of four consecutive terms in a geometric progression is 960. If the fourth term is 8 times as large as the first term, then the smallest number in the geometric progression is

• A. 40
• B. 64
• C. 128
• D. 160
• E. 240

Hint:

When the ratio between two terms is given (e.g., $ar^3 = 8a$), the common ratio $r$ can often be found independently of the first term, greatly simplifying the rest of the calculation.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Let the four terms of the GP be \( a, ar, ar^2, ar^3 \). We use the relationship between the terms and their sum to find the unknown parameters.

Step 2: Detailed Explanation:
Let the terms be \( a, ar, ar^2, ar^3 \).
According to the second condition:
\[ ar^3 = 8a \]
Assuming \( a \neq 0 \), divide by \( a \):
\[ r^3 = 8 \implies r = 2 \]
According to the first condition (Sum of terms):
\[ a + ar + ar^2 + ar^3 = 960 \]
\[ a(1 + r + r^2 + r^3) = 960 \]
Substitute \( r = 2 \):
\[ a(1 + 2 + 4 + 8) = 960 \]
\[ a(15) = 960 \]
\[ a = \frac{960}{15} = 64 \]
Since \( r = 2 \) (which is greater than 1), the terms are increasing.
The smallest term is the first term \( a = 64 \).

Step 3: Final Answer:
The smallest number is 64.