Okay, let's see. I have this problem about a strictly increasing sequence of positive integers where certain subsequences form geometric and arithmetic progressions. The goal is to find the first term, a₁, given that a₁₃ = 2016. Hmm, let me break this down step by step.

First, let's parse the problem. The sequence is strictly increasing, so each term is larger than the previous one. For every positive integer k, two conditions are given:

1. The subsequence a_{2k-1}, a_{2k}, a_{2k+1} is geometric.
2. The subsequence a_{2k}, a_{2k+1}, a_{2k+2} is arithmetic.

So, for each k, the odd-indexed terms form a geometric sequence, and the even-indexed terms form an arithmetic sequence. Wait, not exactly. Let me think. The first condition applies to the terms at positions 2k-1, 2k, 2k+1. So for each k, these three terms form a geometric progression. Similarly, the second condition applies to the terms 2k, 2k+1, 2k+2, forming an arithmetic progression.

Therefore, for each k, the triplet (2k-1, 2k, 2k+1) is geometric, and the triplet (2k, 2k+1, 2k+2) is arithmetic. Since the sequence is strictly increasing, all terms must be positive integers, and each subsequent term is larger than the previous one.

Given that, let's try to model the sequence. Let's denote the terms as follows:

Let me define the terms in terms of k. For the geometric triplets:

For each k, a_{2k} / a_{2k-1} = a_{2k+1} / a_{2k}. So the ratio between consecutive terms in the triplet is constant. Let's denote this common ratio as r_k for the k-th triplet. Similarly, for the arithmetic triplets:

For each k, a_{2k+1} - a_{2k} = a_{2k+2} - a_{2k+1}. The common difference between consecutive terms in the triplet is the same. Let's denote this common difference as d_k for the k-th triplet.

So, from the geometric condition:

a_{2k} = a_{2k-1} * r_k

a_{2k+1} = a_{2k} * r_k = a_{2k-1} * r_k²

From the arithmetic condition:

a_{2k+1} = a_{2k} + d_k

a_{2k+2} = a_{2k+1} + d_k = a_{2k} + 2d_k

But from the geometric condition, we also have:

a_{2k+1} = a_{2k-1} * r_k²

Therefore, equating these two expressions for a_{2k+1}:

a_{2k-1} * r_k² = a_{2k} + d_k

But since a_{2k} = a_{2k-1} * r_k, substitute that in:

a_{2k-1} * r_k² = a_{2k-1} * r_k + d_k

Divide both sides by a_{2k-1} (assuming it's non-zero, which it is since all terms are positive integers):

r_k² = r_k + (d_k / a_{2k-1})

Hmm, this gives a relation between r_k and d_k. Let's note that d_k is the common difference for the arithmetic triplet starting at a_{2k}. Since the sequence is strictly increasing, d_k must be a positive integer. Similarly, r_k must be a rational number greater than 1 (since each term is larger than the previous in the geometric triplet).

But since all terms are integers, the ratios r_k must be integers. Wait, because if a_{2k} = a_{2k-1} * r_k, and both a_{2k} and a_{2k-1} are integers, then r_k must be a rational number. But for the terms to remain integers, r_k must be a rational number such that a_{2k-1} is divisible by the denominator of r_k. However, since we have multiple triplets, this could complicate things. Alternatively, perhaps r_k is an integer. Let's check.

Suppose r_k is an integer. Then, a_{2k} = a_{2k-1} * r_k, and a_{2k+1} = a_{2k} * r_k = a_{2k-1} * r_k². Then, from the arithmetic condition:

a_{2k+1} = a_{2k} + d_k

Which would give d_k = a_{2k+1} - a_{2k} = a_{2k-1} * r_k² - a_{2k-1} * r_k = a_{2k-1} * r_k (r_k - 1)

Since d_k must be a positive integer, and a_{2k-1} is a positive integer, r_k must be an integer greater than 1. So, yes, r_k is an integer ≥ 2. Therefore, all r_k are integers, and d_k is also an integer.

Therefore, we can model the sequence as follows:

For each k ≥ 1:

a_{2k} = a_{2k-1} * r_k

a_{2k+1} = a_{2k} * r_k = a_{2k-1} * r_k²

And for the arithmetic progression:

a_{2k+1} = a_{2k} + d_k

Which gives:

a_{2k-1} * r_k² = a_{2k-1} * r_k + d_k

Therefore:

d_k = a_{2k-1} * r_k (r_k - 1)

Since d_k must be a positive integer, as established earlier.

Therefore, the terms can be generated recursively. Let's try to write out the first few terms in terms of a₁ and r₁, r₂, etc.

Let's denote:

a₁ = a₁

a₂ = a₁ * r₁

a₃ = a₂ * r₁ = a₁ * r₁²

From the arithmetic condition for k=1:

a₃ = a₂ + d₁

Therefore:

a₁ * r₁² = a₁ * r₁ + d₁ => d₁ = a₁ * r₁ (r₁ - 1)

Then, a₄ = a₃ + d₁ = a₁ * r₁² + a₁ * r₁ (r₁ - 1) = a₁ * r₁ (r₁ + (r₁ - 1)) = a₁ * r₁ (2r₁ - 1)

But also, from the geometric condition for k=2:

a₄ = a₃ * r₂ = a₁ * r₁² * r₂

Therefore, equating the two expressions for a₄:

a₁ * r₁ (2r₁ - 1) = a₁ * r₁² * r₂

Divide both sides by a₁ * r₁ (assuming a₁ ≠ 0 and r₁ ≠ 0, which they aren't):

2r₁ - 1 = r₁ * r₂ => r₂ = (2r₁ - 1)/r₁

But r₂ must be an integer, so (2r₁ - 1) must be divisible by r₁. Therefore, r₁ divides (2r₁ - 1). Let's check:

Let r₁ be an integer ≥ 2. Then (2r₁ - 1) divided by r₁ is 2 - 1/r₁. Since r₁ is an integer ≥ 2, 1/r₁ is a fraction less than 1. Therefore, 2 - 1/r₁ is not an integer. But r₂ must be an integer. Therefore, this seems like a contradiction unless r₁ divides (2r₁ - 1). Wait, maybe I made a mistake here.

Wait, let's re-examine. The equation was:

2r₁ - 1 = r₁ * r₂

Therefore, solving for r₂ gives r₂ = (2r₁ - 1)/r₁ = 2 - 1/r₁

But since r₂ must be an integer, 1/r₁ must be an integer. But r₁ is an integer ≥ 2, so 1/r₁ is a fraction. Therefore, 2 - 1/r₁ is not an integer. Therefore, this suggests that there's a contradiction here. That can't be. Therefore, my assumption that r_k is an integer must be wrong? But earlier, I thought that since a_{2k} and a_{2k-1} are integers, r_k must be rational. But maybe r_k is a rational number, not necessarily an integer.

Wait, let's reconsider. Suppose that a_{2k} = a_{2k-1} * r_k, where r_k is a rational number p/q in reduced form. Then, for a_{2k} to be an integer, a_{2k-1} must be divisible by q. Similarly, a_{2k+1} = a_{2k} * r_k = a_{2k-1} * (p/q)^2. For this to be integer, a_{2k-1} must be divisible by q². Therefore, each subsequent term requires the previous term to be divisible by higher powers of q. However, since the sequence is strictly increasing, this might not be sustainable indefinitely, unless q=1. Therefore, perhaps r_k must be integers. Otherwise, the denominators would accumulate, making the terms non-integers.

Therefore, perhaps r_k must indeed be integers. But then, we have the problem that r₂ = 2 - 1/r₁ must be integer, which is only possible if 1/r₁ is integer, which would require r₁=1. But r₁ is the ratio for the first triplet, which is a geometric progression, so r₁ must be greater than 1 (since the sequence is strictly increasing). Therefore, there's a contradiction here.

Hmm, so maybe my initial assumption that the ratios r_k are integers is incorrect. Alternatively, perhaps the ratios are fractions where the denominators divide the numerators in such a way that the terms remain integers.

Wait, let's try an example. Let's suppose that r₁ is 3/2. Then, a₂ = a₁ * 3/2, which must be integer. Therefore, a₁ must be divisible by 2. Let’s say a₁ = 2, then a₂ = 3. Then, a₃ = a₂ * (3/2) = 9/2, which is not an integer. So that doesn't work. Alternatively, if a₁ = 4, then a₂ = 6, a₃ = 9. Then, check the arithmetic condition for k=1: a₃ = a₂ + d₁ => 9 = 6 + d₁ => d₁ = 3. Then, a₄ = a₃ + d₁ = 9 + 3 = 12. Also, from the geometric condition for k=2: a₄ = a₃ * r₂ => 12 = 9 * r₂ => r₂ = 12/9 = 4/3. So, r₂ is 4/3. Then, a₅ = a₄ * r₂ = 12 * 4/3 = 16. Then, check arithmetic condition for k=2: a₅ = a₄ + d₂ => 16 = 12 + d₂ => d₂ = 4. Then, a₆ = 16 + 4 = 20. From geometric condition for k=3: a₆ = a₅ * r₃ => 20 = 16 * r₃ => r₃ = 20/16 = 5/4. Then, a₇ = 20 * 5/4 = 25. Then, arithmetic condition for k=3: a₇ = a₆ + d₃ => 25 = 20 + d₃ => d₃ = 5. Then, a₈ = 25 + 5 = 30. From geometric condition for k=4: a₈ = a₇ * r₄ => 30 = 25 * r₄ => r₄ = 30/25 = 6/5. Then, a₉ = 30 * 6/5 = 36. Arithmetic condition for k=4: a₉ = a₈ + d₄ => 36 = 30 + d₄ => d₄ = 6. Then, a₁₀ = 36 + 6 = 42. Geometric condition for k=5: a₁₀ = a₉ * r₅ => 42 = 36 * r₅ => r₅ = 42/36 = 7/6. Then, a₁₁ = 42 * 7/6 = 49. Arithmetic condition for k=5: a₁₁ = a₁₀ + d₅ => 49 = 42 + d₅ => d₅ = 7. Then, a₁₂ = 49 + 7 = 56. Geometric condition for k=6: a₁₂ = a₁₁ * r₆ => 56 = 49 * r₆ => r₆ = 56/49 = 8/7. Then, a₁₃ = 56 * 8/7 = 64. Wait, but in the problem, a₁₃ is given as 2016. So, in this case, with a₁ = 2, I get a₁₃ = 64. But 2016 is much larger. So, perhaps scaling up?

Wait, but if we can scale the entire sequence by a factor, perhaps all terms would scale. Let's see. Suppose we take the sequence above starting with a₁ = 2, and then multiply each term by a constant factor. However, since ratios like 3/2, 4/3, etc., are involved, scaling by a factor would require the ratios to stay the same. But since the ratios are fractions, unless the scaling factor is a multiple of the denominators, the terms may not remain integers. So, perhaps this approach is not straightforward.

Alternatively, maybe there's a pattern here. Let's note that the ratios r_k are forming a sequence: 3/2, 4/3, 5/4, 6/5, 7/6, etc. Each r_k = (k+2)/(k+1). For example, r₁ = 3/2, r₂ = 4/3, r₃ = 5/4, etc. Then, in that case, the sequence a_n would be a₁, a₁*(3/2), a₁*(3/2)*(4/3) = a₁*2, a₁*2*(5/4) = a₁*(5/2), a₁*(5/2)*(6/5) = a₁*3, etc. Wait, but this seems inconsistent with the arithmetic conditions.

Wait, let's check. If r₁ = 3/2, then a₂ = (3/2)a₁, a₃ = (3/2)a₂ = (9/4)a₁. Then, arithmetic condition: a₃ = a₂ + d₁ => (9/4)a₁ = (3/2)a₁ + d₁ => d₁ = (9/4 - 3/2)a₁ = (9/4 - 6/4)a₁ = (3/4)a₁. Then, a₄ = a₃ + d₁ = (9/4)a₁ + (3/4)a₁ = (12/4)a₁ = 3a₁. Then, geometric condition for k=2: a₄ = a₃ * r₂ => 3a₁ = (9/4)a₁ * r₂ => r₂ = 3a₁ / (9/4)a₁ = 3 / (9/4) = 4/3. Then, a₅ = a₄ * r₂ = 3a₁ * 4/3 = 4a₁. Then, arithmetic condition for k=2: a₅ = a₄ + d₂ => 4a₁ = 3a₁ + d₂ => d₂ = a₁. Then, a₆ = 4a₁ + a₁ = 5a₁. Then, geometric condition for k=3: a₆ = a₅ * r₃ => 5a₁ = 4a₁ * r₃ => r₃ = 5/4. Then, a₇ = 5a₁ * 5/4 = (25/4)a₁. Then, arithmetic condition for k=3: a₇ = a₆ + d₃ => (25/4)a₁ = 5a₁ + d₃ => d₃ = (25/4 - 5)a₁ = (25/4 - 20/4)a₁ = 5/4 a₁. Then, a₈ = (25/4)a₁ + 5/4 a₁ = (30/4)a₁ = (15/2)a₁. Then, geometric condition for k=4: a₈ = a₇ * r₄ => (15/2)a₁ = (25/4)a₁ * r₄ => r₄ = (15/2)/(25/4) = (15/2)*(4/25) = 60/50 = 6/5. Then, a₉ = (15/2)a₁ * 6/5 = (90/10)a₁ = 9a₁. Then, arithmetic condition for k=4: a₉ = a₈ + d₄ => 9a₁ = (15/2)a₁ + d₄ => d₄ = 9a₁ - (15/2)a₁ = (18/2 - 15/2)a₁ = (3/2)a₁. Then, a₁₀ = 9a₁ + (3/2)a₁ = (21/2)a₁. Then, geometric condition for k=5: a₁₀ = a₉ * r₅ => (21/2)a₁ = 9a₁ * r₅ => r₅ = (21/2)/9 = 7/6. Then, a₁₁ = (21/2)a₁ * 7/6 = (147/12)a₁ = (49/4)a₁. Then, arithmetic condition for k=5: a₁₁ = a₁₀ + d₅ => (49/4)a₁ = (21/2)a₁ + d₅ => d₅ = (49/4 - 42/4)a₁ = (7/4)a₁. Then, a₁₂ = (49/4)a₁ + (7/4)a₁ = (56/4)a₁ = 14a₁. Then, geometric condition for k=6: a₁₂ = a₁₁ * r₆ => 14a₁ = (49/4)a₁ * r₆ => r₆ = 14a₁ / (49/4)a₁ = 14 * 4 / 49 = 56/49 = 8/7. Then, a₁₃ = 14a₁ * 8/7 = 16a₁. But in the problem, a₁₃ = 2016, so 16a₁ = 2016 => a₁ = 2016 /16 = 126. Therefore, a₁ = 126.

Wait, so according to this, if the ratios r_k follow the pattern 3/2, 4/3, 5/4, ..., then a₁₃ = 16a₁, so a₁ = 2016 /16 = 126. So, maybe the answer is 126. But let's verify this with the sequence.

Starting with a₁ = 126:

a₁ = 126

a₂ = a₁ * 3/2 = 126 * 3/2 = 189

a₃ = a₂ * 4/3 = 189 * 4/3 = 252

Check arithmetic condition for k=1: a₃ - a₂ = 252 - 189 = 63; so d₁ = 63

Then, a₄ = a₃ + d₁ = 252 + 63 = 315

Check geometric condition for k=2: a₄ / a₃ = 315 / 252 = 5/4 (since 315 = 252 * 5/4). Correct.

Then, arithmetic condition for k=2: a₅ - a₄ = d₂ => a₅ = a₄ + d₂ = 315 + d₂

But also, from geometric condition for k=3: a₅ = a₄ * r₃ = 315 * 5/4 = 393.75. Wait, but 315 * 5/4 is 393.75, which is not an integer. Wait, that's a problem. But earlier, we had a₁ = 126, leading to a₁₃ = 16*126 = 2016. But here, a₅ is not an integer. Wait, that's a contradiction. So, something's wrong here.

Wait, maybe the ratios are not exactly 3/2, 4/3, etc., but it's a different pattern. Wait, in the example earlier, when a₁ = 2, a₅ = 16, which is an integer. But when a₁ = 126, we get a non-integer term. So, perhaps the ratios are not fixed as 3/2, 4/3, etc., but depend on a₁?

Wait, no. Wait, in the earlier example with a₁ = 2, the ratios r_k were indeed 3/2, 4/3, 5/4, etc., leading to a₁₃ = 64, which is 2^6. But when scaling a₁ to 126, which is 2 * 3^2 * 7, the ratios would still be 3/2, 4/3, etc., but then a₁₃ would be 16 * a₁, but 16*126 = 2016, which is given. However, in this case, the terms in between might not be integers. Wait, but in our earlier example with a₁ = 2, all terms were integers. So, why is this discrepancy?

Wait, perhaps the ratios r_k are not exactly 3/2, 4/3, etc., but vary depending on the current term. Wait, let's check the process again. When we derived the ratios, we found that r₂ = (2r₁ - 1)/r₁. For r₂ to be integer, (2r₁ - 1) must be divisible by r₁. Therefore, 2r₁ - 1 = m * r₁, where m is an integer. Then, 2r₁ - m r₁ = 1 => r₁(2 - m) = 1. Since r₁ is a positive integer greater than 1, the only solution is when 2 - m = 1/r₁. But since m is integer, 2 - m must be a fraction with denominator dividing r₁. Therefore, 1/r₁ must be an integer, which only possible if r₁ = 1, which is not allowed. Therefore, this suggests that our initial assumption that the ratios are integers is invalid, leading to a contradiction. Therefore, maybe the ratios are fractions where the denominators divide the numerators in such a way that terms remain integers.

Alternatively, perhaps the ratios r_k are rational numbers, and the denominators of these fractions divide the numerators in a way that each term remains an integer. Let's try to find such a pattern.

Looking back at the initial example:

a₁ = 2

a₂ = 3

a₃ = 4.5, which is not integer. Wait, but in the problem statement, all terms must be integers. Therefore, this suggests that such a ratio as 3/2 would not work because it leads to a non-integer term. Therefore, maybe the ratios have to be integers. But then, as we saw before, the contradiction arises when trying to compute r₂. Therefore, maybe the key is that all ratios r_k must be integers. But then, the problem is unsolvable, which can't be.

Alternatively, perhaps the ratios are not all integers, but each subsequent ratio is determined in such a way that the terms remain integers. For example, maybe the ratios are fractions where the denominators are 1, 2, 3, etc., such that when multiplied, they result in integer terms.

Wait, perhaps the ratios are fractions that are reduced forms where the denominator cancels out with the numerator of the previous term. For example, if the first ratio is 3/2, then a₂ = (3/2)a₁. To make a₂ integer, a₁ must be divisible by 2. Then, a₃ = (4/3)a₂ = (4/3)(3/2)a₁ = 2a₁, which is integer. Then, a₄ = (5/4)a₃ = (5/4)(2a₁) = (5/2)a₁. To make a₄ integer, a₁ must be divisible by 2. So, if a₁ is divisible by 2, then a₄ is integer. Then, a₅ = (6/5)a₄ = (6/5)(5/2)a₁ = 3a₁, which is integer. Then, a₆ = (7/6)a₅ = (7/6)(3a₁) = (7/2)a₁. To make a₆ integer, a₁ must be divisible by 2 again. Then, a₇ = (8/7)a₆ = (8/7)(7/2)a₁ = 4a₁, integer. Then, a₈ = (9/8)a₇ = (9/8)(4a₁) = (9/2)a₁. Again, a₁ must be divisible by 2. Then, a₉ = (10/9)a₈ = (10/9)(9/2)a₁ = 5a₁. Then, a₁₀ = (11/10)a₉ = (11/10)(5a₁) = (11/2)a₁. So, a₁ must be divisible by 2. Then, a₁₁ = (12/11)a₁₀ = (12/11)(11/2)a₁ = 6a₁. Then, a₁₂ = (13/12)a₁₁ = (13/12)(6a₁) = (13/2)a₁. So, a₁ must be divisible by 2. Then, a₁₃ = (14/13)a₁₂ = (14/13)(13/2)a₁ = 7a₁. So, in this case, if a₁ is divisible by 2, then a₁₃ = 7a₁. But in the problem, a₁₃ = 2016. Therefore, 7a₁ = 2016 => a₁ = 2016 /7 = 288. However, in this case, a₁ must be divisible by 2, which 288 is. Let's check if all terms are integers:

a₁ = 288

a₂ = (3/2)*288 = 432

a₃ = (4/3)*432 = 576

a₄ = (5/4)*576 = 720

a₅ = (6/5)*720 = 864

a₆ = (7/6)*864 = 1008

a₇ = (8/7)*1008 = 1152

a₈ = (9/8)*1152 = 1296

a₉ = (10/9)*1296 = 1440

a₁₀ = (11/10)*1440 = 1584

a₁₁ = (12/11)*1584 = 1728

a₁₂ = (13/12)*1728 = 1872

a₁₃ = (14/13)*1872 = 2016

Yes! All terms are integers. Therefore, this works. Therefore, the first term a₁ is 288. But wait, in my initial example, when a₁=2, the a₁₃ was 64, which is 2*32, but in this case, it's 288*7=2016. So, the key is that the ratios are 3/2, 4/3, 5/4, etc., but each time a₁ must be divisible by 2 to make a₂ integer, then a₃ is integer because 4/3 * a₂ is integer if a₂ is divisible by 3, which it is if a₁ is divisible by 2, as 3/2 * a₁. Then, a₄ is 5/4 * a₃. Since a₃ is 4/3 * a₂ = 4/3 * 3/2 * a₁ = 2a₁, which is divisible by 2, so 5/4 * 2a₁ = 5/2 * a₁. Then, a₁ must be divisible by 2 again to make a₄ integer. Then, a₅ is 6/5 * a₄ = 6/5 * 5/2 * a₁ = 3a₁. Then, a₆ is 7/6 * a₅ = 7/6 * 3a₁ = 7/2 * a₁. So, a₁ must be divisible by 2 again. This pattern continues, requiring a₁ to be divisible by 2 each time. Therefore, a₁ must be divisible by 2^6, since the sequence up to a₁₃ requires 6 divisions by 2. Let's check:

Starting with a₁ = 2^6 = 64:

a₂ = 3/2 *64=96

a₃=4/3 *96=128

a₄=5/4 *128=160

a₅=6/5 *160=192

a₆=7/6 *192=224

a₇=8/7 *224=256

a₈=9/8 *256=288

a₉=10/9 *288=320

a₁₀=11/10 *320=352

a₁₁=12/11 *352=384

a₁₂=13/12 *384=416

a₁₃=14/13 *416=448

But 448 ≠ 2016. So, if a₁=64, a₁₃=448. But we need a₁₃=2016, which is 448 * 4.5. Therefore, perhaps the pattern is that each a₁ must be multiplied by a factor to get to 2016. Wait, but 448 is 64*(7). So, 64*7=448. Then, if we need 7a₁=2016, then a₁=288. Let's see:

Start with a₁=288:

a₂=3/2 *288=432

a₃=4/3 *432=576

a₄=5/4 *576=720

a₅=6/5 *720=864

a₆=7/6 *864=1008

a₇=8/7 *1008=1152

a₈=9/8 *1152=1296

a₉=10/9 *1296=1440

a₁₀=11/10 *1440=1584

a₁₁=12/11 *1584=1728

a₁₂=13/12 *1728=1872

a₁₃=14/13 *1872=2016

Yes, so this works. Therefore, the first term a₁ is 288. But wait, why does this work? It seems that each time we have ratios that are fractions where the numerator and denominator are consecutive integers, starting from 3/2, 4/3, etc. Each ratio is (n+1)/n where n starts at 2. Then, each term is multiplied by (n+1)/n, which when applied to the previous term, requires that the previous term is divisible by n to keep the next term integer. Therefore, starting with a₁, which must be divisible by 2 (to make a₂ integer), then a₂ must be divisible by 3 (to make a₃ integer), and so on, up to a term that must be divisible by 12 (to make a₁₃ integer). Therefore, a₁ must be divisible by the least common multiple (LCM) of all the denominators required up to the 12th term. Wait, let's see:

Looking at the ratios:

r₁ = 3/2 (requires a₁ divisible by 2)

r₂ = 4/3 (requires a₂ divisible by 3, which is a₁*3/2 divisible by 3 => a₁ divisible by 2)

r₃ = 5/4 (requires a₃ divisible by 4, which is a₂*4/3 divisible by 4 => a₂ divisible by 3, which is already required)

Similarly, for each subsequent ratio, the denominator is the same as the previous numerator, which is already a multiple. Therefore, the key is that a₁ must be divisible by 2, which then ensures that each subsequent term is divisible by the required denominators. Therefore, the minimal a₁ is 2. But in our example, a₁=2 gave a₁₃=64, which is much smaller than 2016. Therefore, perhaps the general formula is aₙ = a₁ * product of ratios up to that term. Since each ratio is (k+2)/(k+1) for k starting at 1, then the product of these ratios from k=1 to k=12 is (3/2)*(4/3)*(5/4)*...*(14/13) = 14/2 = 7. Therefore, a₁₃ = a₁ * 7. Therefore, if a₁₃ = 2016, then a₁ = 2016 /7 = 288. Therefore, the answer is 288.

Wait, that seems to be the key. Because the product of the ratios from r₁ to r₁₂ is telescoping. Let's verify:

Product of ratios:

r₁ = 3/2

r₂ = 4/3

r₃ = 5/4

...

r₁₂ = 14/13

Multiplying all together:

(3/2)*(4/3)*(5/4)*...*(14/13) = 14/2 = 7

Because all terms cancel out except the first denominator (2) and the last numerator (14). Therefore, a₁₃ = a₁ * 7. Therefore, a₁ = 2016 /7 = 288. Therefore, the answer is 288.

Therefore, the first term is 288.

**Final Answer**
\boxed{288}
Therefore, the first term $a_1$ of the sequence is $\boxed{288}$.
</think>