The first important thing to note is that the way these answer choices are set up, any answer that does not have a at the end - that is, denoting first term of as specified - can be automatically eliminated. The second important thing is realizing that we are given terms that are not consecutive but are two apart, meaning we can use the usual common difference but need to halve it instead of taking it at face value (specifically, .)

The pattern in this sequence is , where  represents the term's place in the sequence. It follows like so:

, our first term.

, our second term.

Then, our third term must be:

. 

Rewrite all three fractions in terms of their least common denominator, which is :

;

 remains as is;

The sequence begins 

Subtract the first term  from the second term  to get the common difference :

Setting  and ,

If this common difference is added a few more times, a pattern emerges:

...

All of the denominators end in 4 or 9, so none of them can be divisible by 20. Therefore, none of the terms will be integers.

Given the first two terms  and , the common difference  of an arithmetic sequence is equal to the difference:

Setting , :

The th term of an arithmetic sequence  can be found by way of the formula

Since we are looking for the first positive number - equivalently, the first number greater than 0:

 for some .

Setting  and , and solving for :

Since  must be a whole number, it follows that the least value of  for which  is ; therefore, the first positive term in the sequence is the twentieth term.

Subtract the first term  from the second term  to get the common difference :

Setting  and ,

The th term of an arithmetic sequence  can be found by way of the formula

Setting , , and  in the formula:

As a fraction, this is 

We start by multiplying 6 times 46, since the first 4 terms are already listed. We then add the product, 276, to the last listed term, 20. This gives us our answer of 296.

All answers in the sequence must end in a 5 or a 0.

Let a₁ represent the first term of the sequence and a_n represent the nth term.

We are told that each term is two greater than the term that precedes it. Thus, we can say that:

a₂ = a₁ + 2

a₃ = a₁ + 2 + 2 = a₁ + 2(2)

a₄ = a₁ + 3(2)

a₅ = a₁ + 4(2)

a_n = a₁ + (n-1)(2)

The problem tells us that the sum of the first five terms is equal to the difference between the fifth and first terms. Let's write an expression for the sum of the first five terms.

sum = a₁ + (a₁ + 2) + (a₁ + 2(2)) + (a₁ + 3(2)) + (a₁ + 4(2))

= 5a₁ + 2 + 4 + 6 + 8

= 5a₁ + 20

Next, we want to write an expression for the difference between the fifth and first terms.

a₅ - a₁ = a₁ + 4(2) – a₁ = 8

Now, we set the two expressions equal and solve for a₁.

5a₁ + 20 = 8

Subtract 20 from both sides.

5a₁ = –12

a₁ = –2.4.

The question ultimately asks us for the tenth term of the sequence. Now, that we  have the first term, we can find the tenth term.

a₁₀ = a₁ + (10 – 1)(2)

a₁₀ = –2.4 + 9(2)

= 15.6

The answer is 15.6 .

Let a₁ be the first term in the sequence. We can use the fact that a_n+1 = (a_n)² – 1 in order to find expressions for the second and third terms of the sequence in terms of a₁.

a₂ = (a₁)² – 1

a₃ = (a₂)² – 1 = ((a₁)² – 1)² – 1

We can use the fact that, in general, (a – b)² = a² – 2ab + b² in order to simplify the expression for a₃.

a₃ = ((a₁)² – 1)² – 1

= (a₁)⁴ – 2(a₁)² + 1 – 1 = (a₁)⁴ – 2(a₁)²

We are told that the third term is equal to the square of the first term. 

a₃ = (a₁)²

We can substitute (a₁)⁴ – 2(a₁)² for a₃.

(a₁)⁴ – 2(a₁)² = (a₁)²

Subtract (a₁)² from both sides.

(a₁)⁴ – 3(a₁)² = 0

Factor out (a₁)² from both terms.

(a₁)² ((a₁)² – 3) = 0

This means that either (a₁)² = 0, or (a₁)² – 3 = 0. 

If (a₁)² = 0, then a₁ must be 0. However, we are told that all the terms of the sequence are positive. Therefore, the first term can't be 0.

Next, let's solve (a₁)² – 3 = 0.

Add 3 to both sides.

(a₁)² = 3

Take the square root of both sides. 

a₁ = ±√3

However, since all the terms are positive, the only possible value for a₁ is √3.

Now, that we know that a₁ = √3, we can find a₂, a₃, and a₄.

a₂ = (a₁)² – 1 = (√3)² – 1 = 3 – 1 = 2

a₃ = (a₂)² – 1 = 2² – 1 = 4 – 1 = 3

a₄ = (a₃)² – 1 = 3² – 1 = 9 – 1 = 8

The question ultimately asks for the product of the a₂, a₃, and a₄, which would be equal to 2(3)(8), or 48.

The answer is 48.

The fourth term is: 3(23) – 1 = 69 – 1 = 68.

The fifth term is: 3(68) – 1 = 204 – 1 = 203.

The sixth term is: 3(203) – 1 = 609 – 1 = 608.