Question

If \( 7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^2}(5 + 2\alpha) + \frac{1}{7^3}(5 + 3\alpha) + \cdots \), then the value of \( \alpha \) is:

• A. \( \frac{6}{7} \)
• B. \( 1 \)
• C. \( \frac{1}{7} \)
• D. \( 6 \)

Hint:

For infinite series with a common ratio, use the sum formula for geometric series and solve for the unknown variable.

Answer 1

The Correct Option is A

An infinite series with a common ratio of \( \frac{1}{7} \) can be expressed as: \[ 7 = 5 + \frac{1}{7} (5 + \alpha) + \frac{1}{7^2} (5 + 2\alpha) + \cdots. \] This is a geometric series. Solving for \( \alpha \) by establishing the sum of the series yields its value.
Final Answer: \( \alpha = \frac{6}{7} \).