The first term of an A.P. is the same as that of a G.P., the common difference of the A.P. and the common ratio of the G.P. are both 2. If the sum of the first five terms of each series be the same, find the `6^(th)` term of each series.

Answer

To solve the problem, we need to find the sixth term of both an Arithmetic Progression (A.P.) and a Geometric Progression (G.P.) given certain conditions. Let's break down the solution step by step.

### Step 1: Define the Variables
Let:
- The first term of both the A.P. and G.P. be \( a \).
- The common difference of the A.P. be \( d = 2 \).
- The common ratio of the G.P. be \( r = 2 \).

### Step 2: Write the Formula for the Sum of the First 5 Terms
The sum of the first \( n \) terms of an A.P. is given by:
\[ S_n = \frac{n}{2} \cdot (2a + (n-1)d) \]
For \( n = 5 \):
\[ S_5 = \frac{5}{2} \cdot (2a + 4d) = \frac{5}{2} \cdot (2a + 4 \cdot 2) = \frac{5}{2} \cdot (2a + 8) \]

The sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = a \cdot \frac{r^n - 1}{r - 1} \]
For \( n = 5 \):
\[ S_5' = a \cdot \frac{2^5 - 1}{2 - 1} = a \cdot (32 - 1) = 31a \]

### Step 3: Set the Sums Equal
Since the sums of the first five terms of both series are equal:
\[ \frac{5}{2} \cdot (2a + 8) = 31a \]

### Step 4: Solve for \( a \)
Multiply both sides by 2 to eliminate the fraction:
\[ 5(2a + 8) = 62a \]
Expanding the left side:
\[ 10a + 40 = 62a \]
Rearranging gives:
\[ 40 = 62a - 10a \]
\[ 40 = 52a \]
Dividing both sides by 52:
\[ a = \frac{40}{52} = \frac{10}{13} \]

### Step 5: Find the Sixth Term of the A.P.
The \( n \)-th term of an A.P. is given by:
\[ A_n = a + (n-1)d \]
For the sixth term:
\[ A_6 = a + (6-1) \cdot d = \frac{10}{13} + 5 \cdot 2 = \frac{10}{13} + 10 \]
Converting 10 to a fraction with a denominator of 13:
\[ A_6 = \frac{10}{13} + \frac{130}{13} = \frac{140}{13} \]

### Step 6: Find the Sixth Term of the G.P.
The \( n \)-th term of a G.P. is given by:
\[ G_n = a \cdot r^{n-1} \]
For the sixth term:
\[ G_6 = a \cdot 2^{6-1} = \frac{10}{13} \cdot 2^5 = \frac{10}{13} \cdot 32 = \frac{320}{13} \]

### Conclusion
The sixth terms are:
- Sixth term of A.P. = \( \frac{140}{13} \)
- Sixth term of G.P. = \( \frac{320}{13} \)