Let `a_(1), a_(2), a_(3)...` be a sequance of positive integers in arithmetic progression with common difference 2. Also let `b_(1),b_(2), b_(3)......` be a sequences of posotive intergers in geometric progression with commo ratio 2. If `a_(1)=b_(1)=c_(2)`. then the number of all possible values of c, for which the equality.
`2(a_(1)+a_(2).+.....+a_(n))=b_(1)+b_(2)+....+b_(n)` holes for same positive integer n, is

Answer

To solve the problem, we need to analyze the sequences and the given equality step by step.

### Step 1: Define the sequences
Let the first term of the arithmetic progression (AP) be \( a_1 = c \) and the common difference be \( d = 2 \). Therefore, the terms of the AP can be expressed as:
\[ a_n = c + (n-1) \cdot 2 = c + 2(n-1) \]

For the geometric progression (GP), let the first term be \( b_1 = c \) and the common ratio be \( r = 2 \). Thus, the terms of the GP can be expressed as:
\[ b_n = c \cdot 2^{n-1} \]

### Step 2: Write the sums of the sequences
The sum of the first \( n \) terms of the AP is given by:
\[ S_a = \frac{n}{2} \cdot (a_1 + a_n) = \frac{n}{2} \cdot \left(c + (c + 2(n-1))\right) = \frac{n}{2} \cdot (2c + 2(n-1)) = n(c + (n-1)) \]

The sum of the first \( n \) terms of the GP is given by:
\[ S_b = b_1 \cdot \frac{r^n - 1}{r - 1} = c \cdot \frac{2^n - 1}{2 - 1} = c(2^n - 1) \]

### Step 3: Set up the equation
According to the problem, we need to satisfy the equation:
\[ 2S_a = S_b \]
Substituting the sums we found:
\[ 2 \cdot n(c + (n-1)) = c(2^n - 1) \]
This simplifies to:
\[ 2n(c + n - 1) = c(2^n - 1) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ 2nc + 2n(n - 1) = c \cdot 2^n - c \]
\[ c \cdot 2^n - 2nc - c = 2n(n - 1) \]
\[ c(2^n - 2n - 1) = 2n(n - 1) \]
Thus, we can express \( c \) as:
\[ c = \frac{2n(n - 1)}{2^n - 2n - 1} \]

### Step 5: Determine conditions for \( c \)
For \( c \) to be a positive integer, the right-hand side must be a positive integer. Therefore, we need:
1. \( 2^n - 2n - 1 > 0 \) (to ensure the denominator is positive).
2. \( 2n(n - 1) \) must be divisible by \( 2^n - 2n - 1 \).

### Step 6: Analyze the inequality
We need to find the smallest \( n \) such that:
\[ 2^n > 2n + 1 \]
Testing values:
- For \( n = 1 \): \( 2^1 = 2 \) and \( 2 \cdot 1 + 1 = 3 \) (not valid)
- For \( n = 2 \): \( 2^2 = 4 \) and \( 2 \cdot 2 + 1 = 5 \) (not valid)
- For \( n = 3 \): \( 2^3 = 8 \) and \( 2 \cdot 3 + 1 = 7 \) (valid)
- For \( n = 4 \): \( 2^4 = 16 \) and \( 2 \cdot 4 + 1 = 9 \) (valid)
- For \( n = 5 \): \( 2^5 = 32 \) and \( 2 \cdot 5 + 1 = 11 \) (valid)
- For \( n = 6 \): \( 2^6 = 64 \) and \( 2 \cdot 6 + 1 = 13 \) (valid)
- For \( n = 7 \): \( 2^7 = 128 \) and \( 2 \cdot 7 + 1 = 15 \) (valid)

### Step 7: Check for integer values of \( c \)
Now we need to check for which values of \( n \) the expression for \( c \) gives a positive integer:
- For \( n = 3 \): \( c = \frac{2 \cdot 3 \cdot 2}{8 - 6} = \frac{12}{2} = 6 \) (valid)
- For \( n = 4 \): \( c = \frac{2 \cdot 4 \cdot 3}{16 - 9} = \frac{24}{7} \) (not an integer)
- For \( n = 5 \): \( c = \frac{2 \cdot 5 \cdot 4}{32 - 11} = \frac{40}{21} \) (not an integer)
- For \( n = 6 \): \( c = \frac{2 \cdot 6 \cdot 5}{64 - 13} = \frac{60}{51} \) (not an integer)
- For \( n = 7 \): \( c = \frac{2 \cdot 7 \cdot 6}{128 - 15} = \frac{84}{113} \) (not an integer)

### Conclusion
The only valid integer value of \( c \) occurs when \( n = 3 \), yielding \( c = 6 \).

Thus, the number of all possible values of \( c \) is:
\[ \boxed{1} \]