Question

In a GP 9, 3, $ \frac{1}{3} $, $ \frac{1}{9} $, … find the 25th term.

• A. \( \frac{1}{3^{24}} \)
• B. \( \frac{9}{3^{24}} \)
• C. \( \frac{9}{3^{25}} \)
• D. \( \frac{1}{3^{25}} \)

Hint:

In a geometric progression, the common ratio \( r \) and the first term \( a \) are key to finding any term in the sequence using the formula \( T_n = a \cdot r^{n-1} \).

Answer 1

The Correct Option is C

We are given a geometric progression (GP) where the first term \( a = 9 \) and the common ratio \( r = \frac{1}{3} \). 
The formula for the \( n \)-th term of a GP is: \[ T_n = a \cdot r^{n-1} \] We are asked to find the 25th term, so \( n = 25 \). 
Substituting the values: \[ T_{25} = 9 \cdot \left( \frac{1}{3} \right)^{24} \] Simplifying: \[ T_{25} = \frac{9}{3^{24}} \] 
Thus, the 25th term is \( \frac{9}{3^{25}} \).