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:
We have a finite geometric progression consisting of four terms.
We use the ratio between terms to find the common ratio \(r\), and then use the sum to find the initial term \(a\).
Step 2: Key Formula or Approach:
The terms are \(a, ar, ar^2, ar^3\).
Condition 1: \(ar^3 = 8a\).
Condition 2: \(a + ar + ar^2 + ar^3 = 960\).
Step 3: Detailed Explanation:
From Condition 1, divide both sides by \(a\) (assuming \(a \neq 0\)):
\[ r^3 = 8 \implies r = 2 \] Substitute \(r = 2\) into Condition 2:
\[ a + a(2) + a(2^2) + a(2^3) = 960 \] \[ a + 2a + 4a + 8a = 960 \] Sum the coefficients:
\[ 15a = 960 \] Solve for \(a\):
\[ a = \frac{960}{15} = 64 \] The progression is 64, 128, 256, 512. Since it is an increasing sequence, the smallest term is the first term.
Step 4: Final Answer:
The smallest number is 64.