Sequences - GRE Subject Test: Math

Card 0 of 19

Question

If a sequence is 9, 3, -3, -9, -15, ...

What is the 10th term of the sequence?

Answer

It will be helpful if you can see that the sequence is changing by -6 to get to each number. The quickest way to solve this is to keep going with the sequence by subtracting six from each number to get to the next number.

9, 3, -3, -9, -15, -21, -27, -33, -39, -45

-45 is the 10th number in the sequence.

Another way to find the 10th term is by using the equation for arithmetic sequences.

where is the term in question, is the first term, is the numbered term, and is the difference of the terms.

To find d subtract the first two terms.

Therefore, the equation becomes

$a_{10}=9+(10-1)(-6)$

$\a_{10}=9+(9)(-6)\a_{10}=9-54\a_{10}=-45$

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Question

The first 3 terms of a geometric sequence are 5,9, and 13. What is the 15th term?

Answer

Step 1: We know it's a geometric sequence, so the difference between each consecutive term is the same. We must find the difference.

Step 2: For every term after the first term, we will add a multiple of to the first term.

Step 3: We can write an equation that helps us find the nth term of the sequence. We add a multiple of 4n to 5; the equation becomes .

Step 4: We need to find the value of n. We want the 15th number and we are given the first number. There are numbers in between, so n=14.

Step 5: Plug in 14 for n and find the 15th term

$15th\ Term=5+4(14)=5+56=61$

The term is

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Question

The first three terms of a arithmetic sequence are . What is the term of this sequence?

Answer

Step 1: We need to find the difference between the terms. To find the difference, subtract the first two terms.

Difference=

Step 2: To find the next term of a arithmetic sequence, we add the difference to the previous term.

Fourth term= term+5
term=. The fourth term is 19.

Step 3: To find the term, we must add a multiple of to the first term. In this problem, we are given the term, so we need to find how many terms are in between the and term.

. This tells me that I have to add fourteen times to get to the term. .

Step 4: Now that we know what the value of n, we can plug it in to the equation:

, where 4 is the starting number, and n represents how many times I need to add 5 to the next term (and up to the 15th term).

Let's plug it in:

.

So, the term in this sequence is .

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Question

What is the next number in the sequence:

Answer

Step 1: Try to find a pattern in the sequence

The pattern is decreasing perfect squares, the difference between each base of the consecutive term is ..

Step 2: Find the next term...

If the base of the next term goes down by , the next term is ..

The next term of this sequence is .

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Question

What is the next term in the sequence:

Answer

Step 1: Try to find the difference between the terms.

Difference between and is .
Difference between and is .
Difference between and is .

Difference between and is .

The difference between consecutive numbers is an odd number, and the difference between the consecutive terms is always ..

The difference between and is

The next number in the sequence is .

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Question

Find the next term in the sequence:

Answer

Step 1: Find the next term in the sequence:

Step 2: Can you recognize the sequence here??

This is the Fibonacci Sequence.. The sum of the previous two terms is equal to the next consecutive term..

The missing term is 13.

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Question

What is the next term in the sequence:

Answer

Step 1: Find the difference between each two consecutive terms...

The difference is always 6 because we have a sequence...

Step 2: Find the next term in the sequence..

The next term is .

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Question

Find the next term in the following sequence:

Answer

Step 1: Calculate the difference between the terms...

$7\rightarrow16$ , add 9
$16\rightarrow25$ , add 9
$25\rightarrow34$ , add 9

Step 2: We found the pattern between the terms..

So, we can add 9 to the term before the to find the value of ..

The missing term (the next term) in the sequence is .

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Question

If the first term of an arithmetic sequence is 2 and the third term is 8, find the $\large \large 16$ th term.

Answer

Step 1: Find the difference between each term...

Subtract the first term from the third term...

$\large 8-2=6$
There are two terms between first and third...Take the answer in step 1 and divide by 2 to get the difference between consecutive terms...

$\large \frac {6}{2}=3$

The common difference is $\large 3$ .

Step 2: Find an equation that describes the sequence....

The equation is $\large 3n+2$ , where $\large n$ represents how many terms I need to calculate and $\large 2$ is the first term...

Step 3: Plug in $\large n$ ...

To find n, we subtract the term that we want from the original term...

So, if we want the $\large 16$ th term and we are given the first term...

Then $\large n=16-1=15$

So, $\large 3(15)+2=45+2=47$

The $\large 16$ th term is $\large 47$ .

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Question

If the first terms of a sequence are ,find the th term.

Answer

Step 1: Find the successive difference rows until we get equal values between every two consecutive numbers.

Step 2: Since we obtain equal values for the successive differences in the second row, hence the $n^{th}$ term of the sequence is a second-degree polynomial.

So, the $n^{th}$ term takes the form of

Step 3: Now, substituting into the formula, we get:

Step 4: Solve the system of linear equations:

So, the $n^{th}$ term takes the form:

Step 5: Plug in into the form...

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Question

Which of the following are not infinite sequences?

$$I. (n!)^{(n!)}, n\ge 0$

$$II. 1\le x\le5; \forall x \in \mathbb{Z}$

$$III. \sqrt[n]{x}; x \in \mathbb{R},n \in \mathbb{N}$

Answer

Step 1: Define what an infinite sequence is...

An infinite sequence is a sequence that is non-terminating.

Step 2: Determine if each sequence above is infinite...

For $(n!)^{(n!)}, n\ge 0$ , the sequence is always infinite because the set of factorials is infinite. Also, the set of values by raising two factorial powers together is also infinite, it never has an ending term.

For $-1\le x \le 5, x\in \mathbb{Z}$ , this sequence is FINITE!

For $\sqrt[n]{x}; x \in \mathbb{R},n \in \mathbb{N}$ , this sequence is FINITE!

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Question

Find the radius of convergence for the power series

$\small \sum_{n=0}^{\infty}\frac{(n!)^2x^n}{(2n)!}$

Answer

We can use the limit

$\small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$

to find the radius of convergence. We have

$\small \small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim \left| \frac{x^{n+1}[(n+1)!]^2}{(2(n+1))!} \frac{(2n!)}{(n!)^2x^n}\right|$

$\small \small \small \small =\lim |x|\left| \frac{(2n)!}{(2n+2)!} \frac{[(n+1)!]^2}{(n!)^2}\right|=\lim |x|\left| \frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|=\frac{|x|}{4}<1$

$\small \implies |x|<4=R$

This means the radius of convergence is $\small R=4$ .

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Question

Determine if the following series is divergent, convergent or neither.

$\sum_{n=1}^{\infty} \frac{n!}{2^{n+1}}$

Answer

In order to figure if

$\sum_{n=1}^{\infty} \frac{n!}{2^{n+1}}$

is convergent, divergent or neither, we need to use the ratio test.

Remember that the ratio test is as follows.

Suppose we have a series $\sum a_n$ . We define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n} \right |$

Then if

, the series is absolutely convergent.

, the series is divergent.

, the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply the ratio test to our problem.

Let

$a_n=\frac{n!}{2^{n+1}}$

and

$a_{n+1}=\frac{{(n+1)!}}{2^{(n+1)+1}}=\frac{(n +1)!}{2^{n+2}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)!}{2^{n+2}}}{\frac{n!}{2^{n+1}}} \right |$

$=\lim_{n\rightarrow \infty}\left | \frac{2^{n+1}(n+1)!}{2^{n+2}n!} \right |$ .

Now lets simplify this expression to

$=\lim_{n\rightarrow \infty}\left | \frac{n+1}{2} \right |$

$=\frac{1}{2}\lim_{n\rightarrow \infty}\left | n+1 \right |$

$=\frac{1}{2}*\infty=\infty$ .

Since $\infty>1$ ,

we have sufficient evidence to conclude that the series is divergent.

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Question

Calculate the sum of the following infinite geometric series:

$\sum_{k=0}^{\infty}17\left(\frac{1}{6}\right)^k$

Answer

This is an infinite geometric series.

The sum of an infinite geometric series can be calculated with the following formula,

$S = \frac{a_{1}}{1-r}$ , where $a_{1}$ is the first value of the summation, and r is the common ratio.

Solution:

Value of $a_{1}$ can be found by setting

$a_{1}=17$

r is the value contained in the exponent

$r=\frac{1}{6}$

$S=\frac{a_{1}}{1-r}$

$S=\frac{17}{1-\frac{1}{6}} = \frac{17*6}{5}$

$S=\frac{102}{5}$

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Question

Determine how many terms need to be added to approximate the following series within $\frac{1}{1000}$ :

$\sum_{n=1}^{\infty} (-1)^n\frac{4}{(n!)^2}$

Answer

This is an alternating series test.

In order to find the terms necessary to approximate the series within $\frac{1}{1000}$ first see if the series is convergent using the alternating series test. If the series converges, find n such that $b_{n+1} < \frac{1}{1000}$

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^n * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$

{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{4}{(n!)^2}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{4}{(n!)^2} =0$

2. { $b_{n}$ } is a decreasing functon, since a factorial never decreases.

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until $b_{n+1} < \frac{1}{1000}$

$b_{4}=\frac{4}{(4!)^2}=\frac{1}{144} > \frac{1}{1000}$

$b_{5}=\frac{4}{(5!)^2}=\frac{1}{3600} < \frac{1}{1000}$

4 needs to be added to approximate the sum within $\frac{1}{1000}$ .

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Question

You have a divergent series $\sum_{n=1}^{\infty} a_{n}$ , and you multiply it by a constant 10. Is the new series $\sum_{n=1}^{\infty}10* a_{n}$ convergent or divergent?

Answer

This is a fundamental property of series.

For any constant c, if $\sum_{n=1}^{\infty} a_{n}$ is convergent then $\sum_{n=1}^{\infty} ca_{n}$ is convergent, and if $\sum_{n=1}^{\infty} a_{n}$ is divergent, $\sum_{n=1}^{\infty} c a_{n}$ is divergent.

$\sum_{n=1}^{\infty} a_{n}$ is divergent in the question, and the constant c is 10 in this case, so $\sum_{n=1}^{\infty} 10*a_{n}$ is also divergent.

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Question

There are 2 series, $\sum_{n=1}^{\infty}a_{n}$ and $\sum_{n=1}^{\infty}b_{n}$ , and they are both convergent. Is $\sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n}$ convergent, divergent, or inconclusive?

Answer

Infinite series can be added and subtracted with each other.

$\sum_{n=1}^{\infty}a_{n} + \sum_{n=1}^{\infty}b_{n} = \sum_{n=1}^{\infty}( a_{n} + b_{n})$

Since the 2 series are convergent, the sum of the convergent infinite series is also convergent.

Note: The starting value, in this case n=1, must be the same before adding infinite series together.

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Question

There are 2 series $\sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n$ and $\sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n$ .

Is the sum of these 2 infinite series convergent, divergent, or inconclusive?

Answer

A way to find out if the sum of the 2 infinite series is convergent or not is to find out whether the individual infinite series are convergent or not.

Test the first series

$\sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n$ .

This is a geometric series with $r = \frac{3}{7} < 1$ .

By the geometric test, this series is convergent.

Test the second series

$\sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n$ .

This is a geometric series with $r = \frac{5}{11} < 1$ .

By the geometric test, this series is convergent.

Since both of the series are convergent, $\sum_{n=1}^{\infty} \left(\frac{3}{7}\right)^n + \sum_{n=1}^{\infty}\left(\frac{5}{11}\right)^n$ is also convergent.

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Question

Evaluate: $\large \large \sum_{x=1}^{10} \frac {1}{n}$ . (Round to 4 places)

Answer

Step 1: Plug in values into the function and add up the fraction:

$\large 1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7}+\frac {1}{8}+\frac {1}{9}+\frac {1}{10}$

Step 2: Find the sum of the fractions....

We can convert the fractions to decimals:

$\large 1+.5+.3333+.25+.2+.1666+.142857+.125+.1111+.1$

$\large =2.92896$

Step 3: Round to $\large 4$ places...

$\large \large 2.92896 \implies 2.9290\implies 2.929$

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