Let a₁, a₂, a₃, ….. be a sequence of positive integers in arithmetic progression with common differe

Q: Wait, why do we reject $n=1$ and $n=2$? The formula gives a result, so why aren't they valid?

That is a fantastic observation. When we plug $n=1$ into our derived expression for $c$, we get $c=0$. However, the problem explicitly states that $a_1$ and $b_1$ are positive integers, meaning $c > 0$. When we plug in $n=2$, we get a negative value for $c$. Since $c$ must be a positive integer, we must reject any $n$ that yields a non-positive result. Always check your boundary conditions against the physical constraints of the problem!

Q: How do I know when to stop checking values of $n$? Is it possible there is a solution for a very large $n$?

This is the core of the problem! We are comparing a polynomial expression ($2n^2 + 1$) with an exponential expression ($2^n$). In the world of functions, exponential growth eventually dominates polynomial growth, no matter how large the coefficients are. Once the exponential function $2^n$ overtakes the polynomial $2n^2 + 1$, it will stay above it forever. By testing values, we see that at $n=7$, the exponential side becomes significantly larger, and it will only grow faster from there. Thus, we can be mathematically certain that no solutions exist for $n \ge 7$.

Q: Why can't we just solve for $n$ directly? Why do we have to test values?

Great question. The equation $c = \frac{2n^2 - 2n}{2^n - 2n - 1}$ is a transcendental-like equation because $n$ appears both in the base and the exponent. There is no simple algebraic method to solve for $n$ in such equations. In competitive exams like JEE Advanced, when you encounter a variable in the exponent and the base, 'bounding' the variable using inequalities is the standard, most elegant strategy. It turns a complex equation into a simple search within a finite range.

Question

Let $a_1, a_2, a_3, \dots$ be a sequence of positive integers in arithmetic progression with common difference 2. Also, let $b_1, b_2, b_3, \dots$ 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 $2(a_1 + a_2 + \dots + a_n) = b_1 + b_2 + \dots + b_n$ holds for some positive integer $n$, is ______

Sigma Percentile · Accepted Answer

### The Dance of Sequences: A Journey into Growth

Imagine you are standing at the starting line of a race. To your left, a runner is moving at a steady, constant pace—this is our Arithmetic Progression (A.P.). To your right, another runner is sprinting with explosive, doubling speed—this is our Geometric Progression (G.P.).

Both start at the same point, $c$. The problem asks us to find when the total distance covered by the steady runner, when doubled, exactly matches the total distance covered by the sprinter. This is not just an algebra problem; it is a story of two different worlds colliding.

### Phase 1: The Mathematical Foundation

For the A.P., the first term is $a_1 = c$ and the common difference is $d = 2$. The sum of the first $n$ terms is given by the classic formula:

$$S_n = \frac{n}{2}[2a_1 + (n-1)d]$$

Substituting our values, we get $S_n = \frac{n}{2}[2c + (n-1)2]$, which simplifies beautifully to:

$$S_n = n(c + n - 1)$$

This is the total distance of our steady runner.

Now, consider the G.P. The first term is $b_1 = c$ and the common ratio is $r = 2$. The sum of the first $n$ terms is $S'_n = b_1 \frac{r^n - 1}{r - 1}$. Substituting $b_1 = c$ and $r = 2$, we get:

$$S'_n = c(2^n - 1)$$

This is the total distance of our explosive sprinter.

### Phase 2: The Algebraic Bridge

The problem gives us a condition: $2(S_n) = S'_n$. Let us set up the equation:

$$2[n(c + n - 1)] = c(2^n - 1)$$

Expanding the left side, we have $2nc + 2n^2 - 2n = c(2^n - 1)$. Our goal is to isolate $c$, the starting value.

Grouping all terms containing $c$ on one side, we get $2n^2 - 2n = c(2^n - 1) - 2nc$. Factoring out $c$, we obtain $2n^2 - 2n = c(2^n - 2n - 1)$. Finally, we isolate $c$:

$$c = \frac{2n^2 - 2n}{2^n - 2n - 1}$$

### Phase 3: The Growth War

We know $c$ must be a positive integer. This implies two things: the denominator $2^n - 2n - 1$ must be positive, and the numerator must be greater than or equal to the denominator (since $c \ge 1$).

Let us analyze the inequality $2n^2 - 2n \ge 2^n - 2n - 1$. Adding $2n$ to both sides, we get the elegant constraint:

$$2n^2 + 1 \ge 2^n$$

We are now comparing a polynomial, $2n^2 + 1$, with an exponential, $2^n$. We know that exponentials grow faster than polynomials. We just need to find the range of $n$ where the polynomial is still holding its own.

Testing the integers:
- For $n=3$: $2(3)^2 + 1 = 19$, and $2^3 = 8$. Since $19 \ge 8$, this is a valid candidate.
- For $n=4$: $2(4)^2 + 1 = 33$, and $2^4 = 16$. Since $33 \ge 16$, this is valid.
- For $n=5$: $2(5)^2 + 1 = 51$, and $2^5 = 32$. Since $51 \ge 32$, this is valid.
- For $n=6$: $2(6)^2 + 1 = 73$, and $2^6 = 64$. Since $73 \ge 64$, this is valid.
- For $n=7$: $2(7)^2 + 1 = 99$, and $2^7 = 128$. Here, $99 < 128$. The exponential has won!

### Phase 4: The Final Verdict

We have narrowed our search to $n \in \{3, 4, 5, 6\}$. Now, we plug these back into our expression for $c$ to see which ones yield an integer:

- For $n=3$: $c = \frac{2(9) - 6}{8 - 6 - 1} = \frac{12}{1} = 12$. This is a valid positive integer!
- For $n=4$: $c = \frac{2(16) - 8}{16 - 8 - 1} = \frac{24}{7}$, which is not an integer.
- For $n=5$: $c = \frac{2(25) - 10}{32 - 10 - 1} = \frac{40}{21}$, not an integer.
- For $n=6$: $c = \frac{2(36) - 12}{64 - 12 - 1} = \frac{60}{51} = \frac{20}{17}$, not an integer.

Out of all possible values, only **$c=12$** works. The beauty of this problem lies in how it forces us to look beyond simple algebra and consider the fundamental nature of growth.

Enter Numerical Value:

Visualized Solution

Given Sequences

Sum Formulas for A.P. and G.P.

Substituting Known Values

Applying the Given Condition

Isolating $c$

Expression for $c$

Checking Small Values of $n$

Bounding the Denominator

Visualizing Polynomial vs Exponential Growth

Finding Valid $n$

Evaluating $c$ for $n = 3$

Evaluating $c$ for $n = 4, 5, 6$

Final Answer

The Sigma Insight: Geometric Progression (G.P.)

The Dance of Sequences

Phase 1

Phase 2

Phase 3

Phase 4

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