Let $$a_{1}, a_{2}, a_{3}$$, … be a sequence of positive integers in arithmetic progression with ommon difference 2. Also, let $$b_{1}, b_{2}, b_{3}$$, … be a sequence of positive integers in geometric progression with common ratio 2. If $$a_{1} = b_{1} = c$$, then the number of all possible values of c, for which the equality

$$2(a_{1} + a_{2} + ... + a_{n}) = (b_{1} + b_{2} + ... + b_{n}$$

hold for some positve integer n, is _______

The arithmetic progression has first term $$a_1=c$$ and common difference $$2$$, so

$$a_k=c+2(k-1) \qquad (k\ge 1).$$

The geometric progression has first term $$b_1=c$$ and common ratio $$2$$, so

$$b_k=c\;2^{\,k-1}\qquad (k\ge 1).$$

Let $$S_a(n)=a_1+a_2+\dots +a_n$$ and $$S_b(n)=b_1+b_2+\dots +b_n$$. For some positive integer $$n$$ we require

$$2S_a(n)=S_b(n).$$

Sum of the arithmetic progression
Using the formula $$S_a(n)=\dfrac{n}{2}\bigl(2a_1+(n-1)d\bigr)$$ with $$a_1=c,\;d=2$$:

$$S_a(n)=\dfrac{n}{2}\bigl(2c+2(n-1)\bigr)=n\,(c+n-1).$$

Sum of the geometric progression
For ratio $$2\neq 1$$, $$S_b(n)=b_1\dfrac{2^{n}-1}{2-1}=c\,(2^{n}-1).$$

Setting up the required equality

$$2S_a(n)=S_b(n) \;\Longrightarrow\; 2\,n\,(c+n-1)=c\,(2^{n}-1).$$

Rearrange to isolate $$c$$:

$$c\bigl(2^{n}-1-2n\bigr)=2n(n-1) \quad -(1)$$

Let $$D(n)=2^{n}-1-2n$$. From $$(1)$$

$$c=\dfrac{2n(n-1)}{D(n)} \quad -(2)$$

We need $$c$$ to be a positive integer, so

1. $$D(n)$$ must be a positive divisor of $$2n(n-1)$$.
2. $$n\ge 2$$, because for $$n=1$$ we would have $$2c=c\Rightarrow c=0$$, which is not positive.

Testing successive values of $$n\ge 2$$

• $$n=2$$: $$D(2)=2^{2}-1-4=4-1-4=-1\;(\lt 0)$$ ⇒ $$c$$ would be negative. Reject.
• $$n=3$$: $$D(3)=2^{3}-1-6=8-1-6=1$$.   Then $$c=\dfrac{2\cdot3\cdot2}{1}=12$$, a positive integer. Accept.
• $$n=4$$: $$D(4)=16-1-8=7$$, but $$2n(n-1)=24$$ is not divisible by $$7$$. Reject.
• $$n=5$$: $$D(5)=32-1-10=21$$, but $$2n(n-1)=40$$ is not divisible by $$21$$. Reject.
• $$n=6$$: $$D(6)=64-1-12=51$$, $$2n(n-1)=60$$ not divisible by $$51$$. Reject.

For $$n\ge 7$$, $$D(n)$$ exceeds $$2n(n-1)$$, so $$c$$ would be less than 1 and hence impossible.

Thus the only valid pair is $$(n,c)=(3,12)$$.

Counting the possible values of $$c$$
Exactly one positive integer value of $$c$$ satisfies the given condition.

Hence, the required number of possible values of $$c$$ is 1.