Question

If the 6th term of a Geometric Progression (GP) is 243 and the 1st term is 32, then what will be the 5th term of the GP?

• A. 162
• B. 81
• C. 108
• D. 72

Answer 1

The Correct Option is A

Step 1: Recall the formula for the \( n \)-th term of a GP.

The \( n \)-th term of a GP is given by:

\[ a_n = a \cdot r^{n-1}, \]

where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.

Step 2: Use the information about the 6th term to find \( r \).

The 6th term is given as \( a_6 = 243 \), and the first term is \( a = 32 \). Using the formula for the \( n \)-th term:

\[ a_6 = a \cdot r^{6-1} = a \cdot r^5. \]

Substitute \( a_6 = 243 \) and \( a = 32 \):

\[ 243 = 32 \cdot r^5. \]

Solve for \( r^5 \):

\[ r^5 = \frac{243}{32}. \]

Take the fifth root of both sides:

\[ r = \left( \frac{243}{32} \right)^{\frac{1}{5}}. \]

Note that \( 243 = 3^5 \) and \( 32 = 2^5 \), so:

\[ r = \frac{3}{2}. \]

Step 3: Find the 5th term (\( a_5 \)).

The formula for the \( n \)-th term is \( a_n = a \cdot r^{n-1} \). For the 5th term:

\[ a_5 = a \cdot r^{5-1} = a \cdot r^4. \]

Substitute \( a = 32 \) and \( r = \frac{3}{2} \):

\[ a_5 = 32 \cdot \left( \frac{3}{2} \right)^4. \]

Compute \( \left( \frac{3}{2} \right)^4 \):

\[ \left( \frac{3}{2} \right)^4 = \frac{3^4}{2^4} = \frac{81}{16}. \]

Thus:

\[ a_5 = 32 \cdot \frac{81}{16} = 2 \cdot 81 = 162. \]

Final Answer: The 5th term of the GP is \( \mathbf{162} \), which corresponds to option \( \mathbf{(1)} \).