Question

If the geometric progressions 162, 54, 18, ..... and \(\frac{2}{81},\frac{2}{27},\frac{2}{9}\),…have their nth term equal, then the value of n is

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

The Correct Option is C

For a geometric progression (G.P.), the general term is given by:

\( a_n = a r^{(n-1)} \)

For the first G.P.: \( 162, 54, 18, \dots \) 

First term: \( a_1 = 162 \)

Common ratio: \( r = \frac{54}{162} = \frac{1}{3} \)

General term: \( a_n = 162 \left(\frac{1}{3}\right)^{(n-1)} \)

For the second G.P.: \( \frac{2}{81}, \frac{2}{27}, \frac{2}{9}, \dots \)

First term: \( a_1 = \frac{2}{81} \)

Common ratio: \( r = \frac{\frac{2}{27}}{\frac{2}{81}} = 3 \)

General term: \( a_n = \frac{2}{81} (3)^{(n-1)} \)

Equating both general terms:

\( 162 \left(\frac{1}{3}\right)^{(n-1)} = \frac{2}{81} (3)^{(n-1)} \)

Multiplying both sides by \( 81 \):

\( 162 \times 81 \left(\frac{1}{3}\right)^{(n-1)} = 2 (3)^{(n-1)} \)

Since \( 162 \times 81 = 13122 \), dividing by 2 gives:

\( 6561 \left(\frac{1}{3}\right)^{(n-1)} = (3)^{(n-1)} \)

Rewriting \( 6561 \) as \( 3^8 \):

\( 3^8 \times 3^{-(n-1)} = 3^{(n-1)} \)

Equating exponents:

\( 8 - (n-1) = n-1 \)

\( 8 = 2(n-1) \)

\( n-1 = 4 \)

\( n = 5 \)

Thus, the required value of n is: 5.

Answer 2

To solve for the value of \(n\) where the nth terms of both given geometric progressions (GPs) are equal, we begin by analyzing each sequence.

The first GP is 162, 54, 18,... with the first term \(a_1 = 162\) and common ratio \(r_1 = \frac{54}{162} = \frac{1}{3}\). 
The nth term of this sequence can be expressed as:

\(T_n^{(1)} = a_1 \cdot r_1^{n-1} = 162 \cdot \left(\frac{1}{3}\right)^{n-1}\)

The second GP is \(\frac{2}{81}, \frac{2}{27}, \frac{2}{9},...\) with the first term \(a_2 = \frac{2}{81}\) and common ratio \(r_2 = \frac{2/27}{2/81} = 3\). 
The nth term of this sequence is:

\(T_n^{(2)} = a_2 \cdot r_2^{n-1} = \frac{2}{81} \cdot 3^{n-1}\)

We need to find \(n\) such that:

\(T_n^{(1)} = T_n^{(2)}\)

Substituting the expressions for the nth terms, we get:

\(162 \cdot \left(\frac{1}{3}\right)^{n-1} = \frac{2}{81} \cdot 3^{n-1}\)

Simplifying both sides, we have:

\(162 \cdot 3^{-(n-1)} = \frac{2}{81} \cdot 3^{n-1}\)

Multiplying both sides by 81 to eliminate the fraction gives:

\(162 \cdot 81 \cdot 3^{-(n-1)} = 2 \cdot 3^{n-1}\)

Recognize that 162 can be expressed as \(162 = 2 \cdot 81\), so the left side becomes:

\(2 \cdot 81^2 \cdot 3^{-(n-1)} = 2 \cdot 3^{n-1}\)

Canceling the factors of 2, and knowing \(81 = 3^4\), we simplify further:

\(3^{8-(n-1)} = 3^{n-1}\)

This leads to:

\(8 - (n-1) = n-1\)

Simplifying the equation:

\(8 - n + 1 = n - 1\)

\(9 - n = n - 1\)

Adding \(n\) to both sides:

\(9 = 2n - 1\)

Adding 1 to both sides gives:

\(10 = 2n\)

Dividing by 2 results in:

\(n = 5\)

Thus, the value of \(n\) is 5.