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High School Math : Finding Terms in a Series

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Example Questions

Example Question #1 : Finding Terms In A Series

Consider the sequence: $2,\ 5,\ 8,\ 11,\ ...$

What is the fifteenth term in the sequence?

Possible Answers:

Correct answer:

Explanation:

The sequence can be described by the equation $\small 2+3(n-1)$ , where is the term in the sequence.

For the 15th term, n=15 .

$\small 2+3(n-1)=2+3(15-1)$

2+3(14)

2+42

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Example Question #31 : Pre Calculus

What are the first three terms in the series?

$\sum_{n=1}^{7}(4n-5)$

Possible Answers:

(3)+(7)+(11)

(1)+(3)+(7)

(-1)+(3)+(7)

(-1)+(3)+(8)

(-5)+(-1)+(3)

Correct answer:

(-1)+(3)+(7)

Explanation:

To find the first three terms, replace with , , and .

(4n-5) = (4(1)-5)=-1

(4n-5) = (4(2)-5)=3

(4n-5) = (4(3)-5)=7

The first three terms are , , and .

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Example Question #1 : Finding Terms In A Series

Find the first three terms in the series.

$\sum_{n=3}^{10}(8-5n)$

Possible Answers:

(3)+(-2)+(-7)

(-2)+(-7)+(-12)

(-7)+(12)+(-17)

(-7)+(-12)+(-17)

(7)+(12)+(17)

Correct answer:

(-7)+(-12)+(-17)

Explanation:

To find the first three terms, replace with , , and .

(8-5n) = (8-5(3))=-7

(8-5n) = (8-5(4))=-12

(8-5n) = (8-5(5))=-17

The first three terms are , , and .

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Example Question #41 : Pre Calculus

Indicate the first three terms of the following series:

$\sum_{x=1}^{7}(4x-5)$

Possible Answers:

(1)+(-3)+(-7)

(-1)+(3)+(7)

(1)+(-3)+(1)

(-1)+(3)+(-1)

(1)+(5)+(9)

Correct answer:

(-1)+(3)+(7)

Explanation:

In the arithmetic series, the first terms can be found by plugging , , and into the equation.

x=1

4(1)-5 = -1

x=2

4(2)-5 = 3

x=3

4(3)-5 = 7

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Example Question #2051 : High School Math

Indicate the first three terms of the following series:

$\sum_{c=3}^{10}(8-5c)$

Possible Answers:

(7)+(-12)+(-17)

(7)+(-12)+(17)

(-7)+(-12)+(-17)

(-7)+(12)+(-17)

(7)+(12)+(17)

Correct answer:

(-7)+(-12)+(-17)

Explanation:

In the arithmetic series, the first terms can be found by plugging in , , and for .

c=3

8-5(3) = -7

c=4

8-5(4) = -12

c=5

8-5(5) = -17

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Example Question #3 : Finding Terms In A Series

Indicate the first three terms of the following series:

$\sum_{m=-2}^{9}(2)^m$

Possible Answers:

$(-\frac{1}{4})+(-\frac{1}{2})+(1)$

$(\frac{1}{4})+(\frac{1}{2})+(0)$

(4)+(2)+(1)

$(-\frac{1}{4})+(-\frac{1}{2})+(-1)$

$(\frac{1}{4})+(\frac{1}{2})+(1)$

Correct answer:

$(\frac{1}{4})+(\frac{1}{2})+(1)$

Explanation:

The first terms can be found by substituting , , and for into the sum formula.

m=-2

$(2)^{-2}=\frac{1}{4}$

m=-1

$(2)^{-1}=\frac{1}{2}$

m=0

$(2)^{0}=1$

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Example Question #4 : Finding Terms In A Series

Indicate the first three terms of the following series.

$\sum_{b=2}^{11} \frac{1}{2}(4)^{b-2}$

Possible Answers:

$\frac{1}{2}\cdot 2\cdot 8$

Not enough information

$\frac{1}{4}\cdot 4\cdot 16$

$\frac{1}{2}+2+8$

$\frac{1}{4}+4+16$

Correct answer:

$\frac{1}{2}+2+8$

Explanation:

The first terms can be found by substituting , , and in for .

b=2

$\frac{1}{2}(4)^{2-2}=\frac{1}{2}$

b=3

$\frac{1}{2}(4)^{3-2}=2$

b=4

$\frac{1}{2}(4)^{4-2}=8$

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Example Question #5 : Finding Terms In A Series

What is the sixth term when $\left (\frac{1}{2}x-2 \right )^{10}$ is expanded?

Possible Answers:

252x^5

-840x^5

840x^6

-252x^5

-840x^6

Correct answer:

-252x^5

Explanation:

We will need to use the Binomial Theorem in order to solve this problem. Consider the expansion of (a+b)^n , where n is an integer. The rth term of this expansion is given by the following formula:

$\binom{n}{r-1}a^{n-r+1}b^{r-1}$ ,

where $\binom{n}{r-1}$ is a combination. In general, if x and y are nonnegative integers such that x > y, then the combination of x and y is defined as follows: $\binom{x}{y}=_xC_y=\frac{x!}{(y)!(x-y)!}$ .

We are asked to find the sixth term of $\left (\frac{1}{2}x-2 \right )^{10}$ , which means that in this case r = 6 and n = 10. Also, we will let $a=\frac{1}{2}x$ and b=-2 . We can now apply the Binomial Theorem to determine the sixth term, which is as follows:

$\binom{10}{6-1}\left (\frac{1}{2}x \right )^{10-6+1}\cdot(-2) ^{6-1}$

$=\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$

Next, let's find the value of $\binom{10}{5}$ . According to the definition of a combination,

$\binom{10}{5}=\frac{10!}{5!5!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5!}{5!5!}$

$=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2}=252$ .

Remember that, if n is a positive integer, then $n!=n\cdot(n-1)\cdot(n-2)\cdot\cdot\cdot3\cdot2\cdot1$ . This is called a factorial.

Let's go back to simplifying $\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}$ .

$\binom{10}{5}\left (\frac{1}{2}x \right )^{5}\cdot(-2) ^{5}=252\cdot\left (\frac{1}{2}x \right )^{5}\cdot-32$

$=252\cdot\left (\frac{1}{2} \right )^5\cdot x^5\cdot (-32) =252\cdot \frac{1}{32}\cdot-3 2\cdot x^5$

=-252x^5

The answer is -252x^5 .

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High School Math : Finding Terms in a Series

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Finding Terms In A Series

Example Question #31 : Pre Calculus

Example Question #1 : Finding Terms In A Series

Example Question #41 : Pre Calculus

Example Question #2051 : High School Math

Example Question #3 : Finding Terms In A Series

Example Question #4 : Finding Terms In A Series

Example Question #5 : Finding Terms In A Series

All High School Math Resources

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