Question

The sixth term in the sequence \( 3, 1, \frac{1}{3}, \dots \) is:

• A. \( \frac{1}{27} \)
• B. \( \frac{1}{9} \)
• C. \( \frac{1}{81} \)
• D. \( \frac{1}{17} \)
• E. \( \frac{1}{7} \)

Hint:

When the common ratio is a fraction less than 1, the terms will decrease rapidly. You can often verify your answer by simply writing out the terms manually for small \( n \): \( 3, 1, 1/3, 1/9, 1/27, 1/81 \).

Answer 1

The Correct Option is C

Concept: By examining the ratio between consecutive terms, we can determine the nature of the sequence. In this case, \( \frac{1}{3} = \frac{1}{3} \) and \( \frac{1/3}{1} = \frac{1}{3} \). Since the ratio is constant, this is a Geometric Progression (G.P.). The formula for the \( n^{th} \) term of a G.P. is \( a_n = ar^{n-1} \).

Step 1: Identifying the sequence parameters.
From the sequence \( 3, 1, \frac{1}{3}, \dots \): - The first term \( a = 3 \). - The common ratio \( r = \frac{1}{3} \). - We need to find the sixth term, so \( n = 6 \).

Step 2: Calculating the sixth term.
Substitute the values into the G.P. formula: \[ a_6 = 3 \times \left( \frac{1}{3} \right)^{6-1} \] \[ a_6 = 3 \times \left( \frac{1}{3} \right)^5 \] \[ a_6 = 3 \times \frac{1}{243} \] \[ a_6 = \frac{1}{81} \]