Let's call the first term in the sequence a₁ and the nth term a_n. 

We are told that each term is r times larger than the one before it. Thus, we can find the next term in the sequence by multiplying by r. 

a₁ = a₁

a₂ = r(a₁)

a₃ = r(a₂) = r(r(a₁)) = r²(a₁)

a₄ = r(a₃) = r(r²(a₁)) = r³(a₁)

a_n = r^(n–1)a₁

We can use this information to find r.  

The problem gives us the value of the third and the sixth terms. 

a₃ = r²(a₁) = 12

a₆ = r⁵(a₁) = 96

Let's solve for a₁ in terms of r and a₃. 

a₁ = 12/(r²)

Let's then solve for a₁ in terms of r and a₆.

a₁ = 96/(r⁵)

Now, we can set both values equal and solve for r.

12/(r²) = 96/(r⁵)

Multiply both sides by r⁵ to get rid of the fraction.

12r⁵/r² = 96

Apply the property of exponents which states that a^b/a^c = a^b–c.

12r³ = 96

Divide by 12 on both sides.

r³ = 8

Take the cube root of both sides.

r = 2

This means that each term is two times larger than the one before it, or that each term is one half as large as the one after it. 

a₂ must equal a₃ divided by 2, which equals ¹²/₂ = 6.

a₁ must equal a₂ divided by 2, which equals ⁶/₂ = 3.

Here are the first eight terms of the sequence:

3, 6, 12, 24, 48, 96, 192, 384

The question asks us to find the sum of all the terms less than 250. Only the first seven terms are less than 250. Thus the sum is equal to the following:

sum = 3 + 6 + 12 + 24 + 48 + 96 + 192 = 381

Given terms  of an arithmetic sequence, , the common difference  is 

Setting , the common difference is found to be

The  term of the sequence  is 

Setting , this term is found to be

It is not necessary to determine the common difference or any other terms to answer this question. If we let  be the common difference, then

and 

, or 

The sum of the second and ninth terms is given by the expression

Given terms  of an arithmetic sequence, , the common difference  is 

.

Set :

Natural numbers are defined as whole numbers 1 and above. II, IV, V are not natural numbers.

All of the values in the sequence must be a multiple of 3. All answers are multiples of 3 except 461 so 461 cannot be part of the sequence. 

The fourth term of this sequence will be 5m + 2m = 7m. If we add up the first four terms, we get m + 3m + 5m + 7m = 4m + 12m = 16m. Since m is an integer, the sum of the first four terms, 16m, will have a factor of 16. Looking at the answer choices, 60 is the only answer where 16 is not a factor, so that is the correct choice.

Let d represent the common difference between consecutive terms.

Let a_n denote the nth term in the sequence. 

In order to get from the tenth term to the twentieth term in the sequence, we must add d ten times.

Thus a_{20 =} a₁₀ + 10d

20 = 40 + 10d

d = -2

In order to get from the first term to the tenth term, we must add d nine times.

Thus a_{10 =} a_{1 +} 9d

40 = a₁ + 9(-2)

The first term of the sequence must be 58.

Our sequence looks like this: 58,56,54,52,50…

We are asked to find the nth term such that the sum of the first n numbers in the sequence equals 0.

58 + 56 + 54 + …. a_{n =} 0

Eventually our sequence will reach zero, after which the terms will become the negative values of previous terms in the sequence.

58 + 56 + 54 + … 6 + 4 + 2 + 0 + -2 + -4 + -6 +….-54 + -56 + -58 = 0

The sum of the term that equals -2 and the term that equals 2 will be zero. The sum of the term that equals -4 and the term that equals 4 will also be zero, and so on.

So, once we add -58 to all of the previous numbers that have been added before, all of the positive terms will cancel, and we will have a sum of zero. Thus, we need to find what number -58 is in our sequence.

It is helpful to remember that a_n = a₁ + d(n-1), because we must add d to a₁ exactly n-1 times in order to give us a_n. For example, a₅ = a₁ + 4d, because if we add d four times to the first term, we will get the fifth term. We can use this formula to find n.

-58 = a_n = a₁ + d(n-1)

-58 = 58 + (-2)(n-1)

n = 59

We can see how the sequence begins by writing out the first few terms:

1, –2, 4, –8, 16, –32, 64, –128.

Notice that every other term (of which there are exactly 50/2 = 25) is negative and therefore less than 25. Also notice that after the fourth term, every term is greater in absolute value than 5, so we just have to find the number of positive terms before the fourth term that are less than 5 and add that number to 25 (the number of negative terms in the first 50 terms).

Of the first four terms, there are only two that are less than 5 (i.e. 1 and 4), so we include these two numbers in our count: 25 negative numbers plus an additional 2 positive numbers are less than 5, so 27 of the first 50 terms of the sequence are less than 5.

Look for cancellations to simplify.  The sum of all consecutive integers from  to  is equal to . Therefore, we must go a little farther.  , so the last number in the sequence in .  That gives us  negative integers,  positive integers, and don't forget zero! .