The nth term of a progression is `3^(n+1)`. Show that it is a G.P. Also find its 5th term.

Explanation

To determine if the sequence defined by the nth term \( a_n = 3^{n+1} \) is a geometric progression (G.P.) and to find its 5th term, we can follow these steps:

### Step 1: Identify the nth term
The nth term of the progression is given as:
\[ a_n = 3^{n+1} \]

### Step 2: Calculate the first few terms
To analyze the progression, we will calculate the first few terms:
- For \( n = 1 \):
\[ a_1 = 3^{1+1} = 3^2 = 9 \]
- For \( n = 2 \):
\[ a_2 = 3^{2+1} = 3^3 = 27 \]
- For \( n = 3 \):
\[ a_3 = 3^{3+1} = 3^4 = 81 \]
- For \( n = 4 \):
\[ a_4 = 3^{4+1} = 3^5 = 243 \]

### Step 3: Check for a common ratio
To show that this sequence is a G.P., we need to check if the ratio of consecutive terms is constant. We can calculate the common ratio \( r \):
\[ r = \frac{a_2}{a_1} = \frac{27}{9} = 3 \]
\[ r = \frac{a_3}{a_2} = \frac{81}{27} = 3 \]
\[ r = \frac{a_4}{a_3} = \frac{243}{81} = 3 \]

Since the common ratio \( r \) is the same (3) for all pairs of consecutive terms, we conclude that the sequence is indeed a geometric progression.

### Step 4: Find the 5th term
To find the 5th term, we substitute \( n = 5 \) into the nth term formula:
\[ a_5 = 3^{5+1} = 3^6 \]
Calculating \( 3^6 \):
\[ 3^6 = 729 \]

### Conclusion
Thus, we have shown that the sequence is a geometric progression with a common ratio of 3, and the 5th term is:
\[ \text{5th term} = 729 \]