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 a₁represent the first term of the sequence and a_nrepresent 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

a_3.= 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 .

Leta₁be the first term in the sequence. We can use the fact thata_n+1= (a_n)²– 1 in order to find expressions for the second and third terms of the sequence in terms ofa₁.

a₂= (a₁)²– 1

a_3.= (a₂)²– 1 = ((a₁)²– 1)²– 1

We can use the fact that, in general, (a –b)²=a²– 2ab+b²in order to simplify the expression fora_3..

a_3.= ((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.

a_3.= (a₁)²

We can substitute (a₁)⁴– 2(a₁)²fora_3..

(a₁)⁴– 2(a₁)²= (a₁)²

Subtract (a₁)²from both sides.

(a₁)⁴– 3(a₁)²= 0

Factor out (a₁)²from both terms.

(a₁)²((a₁)²– 3) = 0

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

If (a₁)²= 0, thena₁must 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 fora₁is √3.

Now, that we know thata₁= √3, we can finda₂,a_3., anda₄.

a₂= (a₁)²– 1 = (√3)²– 1 = 3 – 1 = 2

a_3.= (a₂)²– 1 = 2²– 1 = 4 – 1 = 3

a₄= (a_3.)²– 1 = 3²– 1 = 9 – 1 = 8

The question ultimately asks for the product of thea₂,a_3., anda₄, 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.

Each number in the sequence in 7 more than the number preceding it.

The equation for the terms in an arithmetic sequence is a_n= a₁+ d(n-1), where d is the difference.

The formula for the terms in this sequence is therefore a_n= 2 + 7(n-1).

Plug in 8 for n to find the 8^thterm:

a₈= 2 + 7(8-1) = 51

The difference between each term can be found through subtraction. For example the difference between the first and the second term can be found as follows:

One can check and see that this is the case for the other given numbers in the sequence as well.

In order to find the seventh term, expand the sequence by adding 14 to the last given number (4th number) and all of the following numbers until the 7th number in the sequence is reached.

This gives the sequence:

As seen above the seventh number in the sequence is 87 and the correct answer.

The purpose of this question is to understand the patterns of sequences.

First, an equation for theterm in the sequence must be determined ().

This is true because

will create,

will create,

will create,

will create.

Then, the eqution must be applied to find the specified term. For the tenth term, the expressionmust be evaluated, yielding 103.

From the given information, we know, which means each consecutive difference is 3.

Subtract the first termfrom the second termto get the common difference:

Settingand

Theth term of an arithmetic sequencecan be found by way of the formula

Setting,, andin the formula: