Let `a_1,a_2,a_3 , . . . a_n` are in A.P and `sum_(r=1)^oo a^r/2^r=4`, then `4a_2` is equal to

Answer

To solve the problem, we need to find the value of \(4a_2\) given that the sequence \(a_1, a_2, a_3, \ldots, a_n\) is in arithmetic progression (A.P.) and the sum

\[ \sum_{r=1}^{\infty} \frac{a^r}{2^r} = 4. \]

### Step-by-step Solution:

1. **Understanding the Sum**:
The sum can be rewritten as:
\[ \sum_{r=1}^{\infty} \frac{a_r}{2^r} = \frac{a_1}{2^1} + \frac{a_2}{2^2} + \frac{a_3}{2^3} + \ldots \]
This is an infinite series where \(a_r\) are the terms of the A.P.

2. **Expressing \(a_r\)**:
Since \(a_1, a_2, a_3, \ldots\) are in A.P., we can express \(a_r\) as:
\[ a_r = a_1 + (r-1)d, \]
where \(d\) is the common difference.

3. **Substituting \(a_r\) into the Sum**:
Substitute \(a_r\) into the sum:
\[ \sum_{r=1}^{\infty} \frac{a_1 + (r-1)d}{2^r} = 4. \]
This can be split into two separate sums:
\[ \sum_{r=1}^{\infty} \frac{a_1}{2^r} + \sum_{r=1}^{\infty} \frac{(r-1)d}{2^r} = 4. \]

4. **Calculating the First Sum**:
The first sum is a geometric series:
\[ \sum_{r=1}^{\infty} \frac{a_1}{2^r} = a_1 \sum_{r=1}^{\infty} \left(\frac{1}{2}\right)^r = a_1 \cdot \frac{\frac{1}{2}}{1 - \frac{1}{2}} = a_1. \]

5. **Calculating the Second Sum**:
The second sum can be calculated using the formula for the sum of an infinite series:
\[ \sum_{r=1}^{\infty} \frac{(r-1)d}{2^r} = d \sum_{r=1}^{\infty} \frac{(r-1)}{2^r}. \]
The sum \(\sum_{r=1}^{\infty} \frac{(r-1)}{2^r}\) can be derived from the formula for the sum of an infinite series:
\[ \sum_{r=1}^{\infty} \frac{r}{x^r} = \frac{x}{(x-1)^2}. \]
For \(x = 2\):
\[ \sum_{r=1}^{\infty} \frac{r}{2^r} = \frac{2}{(2-1)^2} = 2. \]
Thus,
\[ \sum_{r=1}^{\infty} \frac{(r-1)}{2^r} = 2 - 1 = 1. \]
Therefore, the second sum becomes:
\[ d \cdot 1 = d. \]

6. **Combining the Results**:
Now we have:
\[ a_1 + d = 4. \]

7. **Finding \(4a_2\)**:
We know that:
\[ a_2 = a_1 + d. \]
Therefore,
\[ 4a_2 = 4(a_1 + d) = 4 \cdot 4 = 16. \]

### Final Answer:
\[ 4a_2 = 16. \]