Question

A geometric progression consists of an even number of termsIf the sum of all the terms is five times the sum of the terms occupying the odd places, then the common ratio of the geometric progression is

• A. \( r = 4 \)
• B. \( r = 3 \)
• C. \( r = 6 \)
• D. \( r = 2 \)

Hint:

Odd terms of a GP form another GP with common ratio \( r^2 \)Use this to simplify such problems quickly.

Answer 1

The Correct Option is A

Step 1: Assume the GP.
Let GP be:
\[ a, ar, ar^2, ar^3, \ldots \]
Total number of terms = \( 2n \) (even).

Step 2: Sum of all terms.
\[ S = a + ar + ar^2 + \cdots + ar^{2n-1} \]
\[ S = \frac{a(1 - r^{2n})}{1 - r} \]

Step 3: Sum of odd place terms.
Odd terms:
\[ a, ar^2, ar^4, \ldots, ar^{2n-2} \]
This is a GP with ratio \( r^2 \).
\[ S_{odd} = \frac{a(1 - r^{2n})}{1 - r^2} \]

Step 4: Use given condition.
\[ S = 5 S_{odd} \]
\[ \frac{a(1 - r^{2n})}{1 - r} = 5 \cdot \frac{a(1 - r^{2n})}{1 - r^2} \]

Step 5: Cancel common terms.
\[ \frac{1}{1 - r} = \frac{5}{1 - r^2} \]

Step 6: Simplify equation.
\[ 1 - r^2 = (1 - r)(1 + r) \]
So:
\[ \frac{1}{1 - r} = \frac{5}{(1 - r)(1 + r)} \]
Cancel \( (1-r) \):
\[ 1 = \frac{5}{1 + r} \]
\[ 1 + r = 5 \]
\[ r = 4 \]

Step 7: Final conclusion.
\[ \boxed{r = 4} \]