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 $\displaystyle -28$ to $\displaystyle 28$ is equal to $\displaystyle 0$. Therefore, we must go a little farther.  $\displaystyle 0+29+30=59$, so the last number in the sequence in $\displaystyle 30$.  That gives us $\displaystyle 28$ negative integers, $\displaystyle 30$ positive integers, and don't forget zero! $\displaystyle 28+30+1=59$.

If Brad can walk 3600 feet in 10 minutes, then he can walk 3600/10 = 360 feet per minute, and 360/60 = 6 feet per second.

There are 3 feet in a yard, so Brad can walk 6/3 = 2 yards per second, or 2 x 10 = 20 yards in 10 seconds.

The first important thing to note is that the way these answer choices are set up, any answer that does not have a $\displaystyle + 2$ at the end - that is, denoting first term of $\displaystyle 2$ 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, $\displaystyle \frac{6}{2} = 3$.)

The pattern in this sequence is $\displaystyle 5^n-1$, where $\displaystyle n$ represents the term's place in the sequence. It follows like so:

$\displaystyle 5^n-1$

$\displaystyle 5^1-1=4$, our first term.

$\displaystyle 5^2-1=24$, our second term.

Then, our third term must be:

$\displaystyle 5^3-1 = 124$. 

Rewrite all three fractions in terms of their least common denominator, which is $\displaystyle LCM (5, 10, 20 ) = 20$:

$\displaystyle \frac{1}{5} =\frac{1 \cdot 4}{5 \cdot 4} = \frac{4}{20}$;

$\displaystyle \frac{9}{20}$ remains as is;

$\displaystyle \frac{7}{10} = \frac{7 \cdot 2}{10 \cdot 2} = \frac{14}{20}$

The sequence begins 

$\displaystyle \frac{4}{20},\frac{9}{20}, \frac{14}{20}...$

Subtract the first term $\displaystyle a_{1}$ from the second term $\displaystyle a_{2}$ to get the common difference $\displaystyle d$:

$\displaystyle d = a_{2} - a_{1}$

Setting $\displaystyle a_{2} = \frac{9}{20}$ and $\displaystyle a_{1} = \frac{4}{20}$,

$\displaystyle d = \frac{9}{20} -\frac{4}{20} = \frac{5}{20}$

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

$\displaystyle \frac{14}{20} + \frac{5}{20} = \frac{19}{20}$

$\displaystyle \frac{19}{20} + \frac{5}{20} = \frac{24}{20}$...

$\displaystyle \frac{4}{20},\frac{9}{20}, \frac{14}{20} ,\frac{19}{20}, \frac{24}{20} ...$

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 $\displaystyle a_{1}$ and $\displaystyle a_{2}$, the common difference $\displaystyle d$ of an arithmetic sequence is equal to the difference:

$\displaystyle d = a_{2} - a_{1}$

Setting $\displaystyle a_{1} = -129$, $\displaystyle a_{2} =-122$:

$\displaystyle d = -122 - (-129) = -122 + 129 = 7$

The $\displaystyle n$th term of an arithmetic sequence $\displaystyle a_{n}$ can be found by way of the formula

$\displaystyle a_{n} = a_{1}+ (n-1) d$

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

$\displaystyle a_{n} = a_{1}+ (n-1) d > 0$ for some $\displaystyle n$.

Setting $\displaystyle a_{1} = -129$ and $\displaystyle d = 7$, and solving for $\displaystyle n$:

$\displaystyle -129 + (n-1) 7 > 0$

$\displaystyle -129 + 7n- 7 > 0$

$\displaystyle 7n- 136 > 0$

$\displaystyle 7n > 136$

$\displaystyle \frac{7n}{7} > \frac{136}{7}$

$\displaystyle n > 19\frac{3}{7}$

Since $\displaystyle n$ must be a whole number, it follows that the least value of $\displaystyle n$ for which $\displaystyle a_{n}> 0$ is $\displaystyle n = 20$; therefore, the first positive term in the sequence is the twentieth term.