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Calculus 2 : Arithmetic and Geometric Series

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Example Questions

Example Question #2971 : Calculus Ii

Calculate the sum of the following series: $\sum^{400}_{k=20}2k$

Possible Answers:

160000

160220

160020

160040

Correct answer:

160020

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

When k is not equal to 1, split the sums up into 2 piece and take the difference.

Find $\sum^{n}_{k=1}2k - \sum^{k-1}_{k=1}2k$

Solution:

$a_{k}=2k$

$a_{1} = 2$

$a_{19} = 2*19 = 38$

$a_{400} = 2*400 = 800$

$\sum^{400}_{k=1}2k= \frac{n}{2}(a_{1}+a_{400}) = \frac{400}{2}(2+800) = 200*802 = 160400$

$\sum^{19}_{k=1}2k= \frac{n}{2}(a_{1}+a_{19})= \frac{19}{2}(2+38) = 380$

$\sum^{400}_{k=20}2k = 160400 - 380 = 160020$

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Example Question #2972 : Calculus Ii

Calculate the sum of the following series: $\sum^{300}_{k=100}5k-60$

Possible Answers:

188040

188940

189000

188900

Correct answer:

188940

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

When k is not equal to 1, split the sums up into 2 piece and take the difference.

Find $\sum^{n}_{k=1}5k-60 - \sum^{k-1}_{k=1}5k-60$

Solution:

$a_{k}=5k-60$

$a_{1} = 5*1-60 = -55$

$a_{99} = 5*99-60=435$

$a_{300} = 300*5-60 = 1500-60 = 1440$

$\sum^{300}_{k=1}5k-60= \frac{n}{2}(a_{1}+a_{300}) = \frac{300}{2}(-55 + 1440)=150*1385=207750$

$\sum^{99}_{k=1}5k-60= \frac{n}{2}(a_{1}+a_{99}) = \frac{99}{2}(-55+435)=\frac{99}{2}380 = 18810$

$\sum^{300}_{k=100}5k-60 = 207750 - 18810 = 188940$

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Example Question #191 : Series In Calculus

Calculate the sum of the following series: $\sum^{100}_{k=60}15k-30$

Possible Answers:

47900

47910

47970

47930

Correct answer:

47970

Explanation:

This is an arithmetic series.

Its general form is $\sum^{n}_{k=1}a_k$

To calculate the sum of very large series such as these, use the formula $\frac{n}{2}(a_1 + a_n)$

This works because you are taking the average of the largest and smallest terms, and then multiplying them by n, which is the same as calculating the sum total.

When k is not equal to 1, split the sums up into 2 piece and take the difference.

Find $\sum^{n}_{k=1}15k-30 - \sum^{k-1}_{k=1}15k-30$

Solution:

$a_{k}=15k-30$

$a_{1}=15*1-30 = -15$

$a_{59}=15*59 -30 = 885-30 = 855$

$a_{100} = 100*15-30 = 1500-30 = 1470$

$\sum^{100}_{k=1}15k-30= \frac{n}{2}(a_{1}+a_{100})=\frac{100}{2}(-15+1470)=1455*50=72750$

$\sum^{59}_{k=1}15k-30= \frac{n}{2}(a_{59}+a_{1}) = \frac{59}{2}(855-15)=59*420=24780$

$\sum^{100}_{k=60}15k-30= 72750-24780=47970$

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Example Question #192 : Series In Calculus

Find the value of the 700th term, or $a_{700}$ , in the following arithmetic series:

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+... =7+15+23+31+39+...$

Possible Answers:

5598

5599

6002

6000

Correct answer:

5599

Explanation:

3 pieces of information are needed to find the value of a specific term

First, find the first value, $a_{1}$ .

Second, the value of n, or the total number of terms in the series.

Finally, d, or the common difference, which can be found by calculating $a_{2} -a{1}$ .

With these pieces of information, find the value of the last with the following formula:

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

Solution:

$a_{1}=7$

n=700

$d = a_{2}-a_{1}=15-7=8$

$a_{700} = a_{1} + (n-1)*d = 7 + (700-1)*8=7+5592=5599$

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Example Question #2975 : Calculus Ii

Find the value of the 400th term, or $a_{400}$ , in the following arithmetic series:

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+...=3+9+15+21+27+...$

Possible Answers:

2402

2399

2400

2397

Correct answer:

2397

Explanation:

3 pieces of information are needed to find the value of a specific term.

First, find the first value, . $a_{1}$

Second, the value of n, or the total number of terms in the series.

Finally, d, or the common difference, which can be found by calculating $a_{2} -a{1}$ .

With these pieces of information, find the value of the last with the following formula:

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

Solution:

$a_{1}=3$

n=400

$d = a_{2}-a_{1}=9-3=6$

$a_{400} = a_{1} + (n-1)*d=3+(400-1)*6 = 3+2394=2397$

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Example Question #2981 : Calculus Ii

Find the value of the 1200th term, or $a_{1200}$ , in the following arithmetic series:

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+...=101+124+147+170+193+...$

Possible Answers:

27668

27681

27680

27678

Correct answer:

27678

Explanation:

3 pieces of information are needed to find the value of a specific term.

First, find the first value, $a_{1}$ .

Second, the value of n, or the total number of terms in the series.

Finally, d, or the common difference, which can be found by calculating $a_{2} -a{1}$ .

With these pieces of information, find the value of the last with the following formula:

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

Solution:

$a_{1}=101$

n=1200

$d = a_{2}-a_{1}=124-101=23$

$a_{1200} = a_{1} + (n-1)*d=101+(1200-1)*23=101+27577=27678$

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Example Question #2981 : Calculus Ii

Find the value of the 300th term, or $a_{300}$ , in the following arithmetic series:

$a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+...=40+57+74+91+108+...$

Possible Answers:

5123

5125

5120

5115

Correct answer:

5123

Explanation:

3 pieces of information are needed to find the value of a specific term.

First, find the first value, $a_{1}$ .

Second, the value of n, or the total number of terms in the series.

Finally, d, or the common difference, which can be found by calculating $a_{2} -a{1}$ .

With these pieces of information, find the value of the last with the following formula:

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

Solution:

$a_{1}=40$

n=300

$d = a_{2}-a_{1}=57-40=17$

$a_{300} = a_{1} + (n-1)*d=40+(300-1)*17=40+5083=5123$

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Example Question #21 : Types Of Series

Calculate the sum of a geometric series with the following values: n=15 , $a_{1}=4$ , $r=\frac{1}{3}$ . Round the answer to the nearest integer.

Possible Answers:

Correct answer:

Explanation:

This is a geometric series.

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

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$S_{15}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{15}=\frac{4*(1-(\frac{1}{3})^{15})}{1-\frac{1}{3}}$

$S_{15}=\frac{4*3(1-\frac{1}{3^{15}})}{2}$

$S_{15}=6*(1-\frac{1}{3^{15}})$

$S_{15}=5.999...$

Rounding, $S_{15}=6$

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Example Question #2983 : Calculus Ii

Calculate the sum, rounded to the nearest integer, of the first 20 terms of the following geometric series: 4+6+9+13.5+20.25...

Possible Answers:

26594

26597

26595

26596

Correct answer:

26594

Explanation:

This is a geometric series.

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

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

We have $a_{1}$ and n and we just need to find r before calculating the sum.

Solution:

$a_{1}=4$

$r=\frac{a_{2}}{a_{1}}$

$r=\frac{6}{4} = 1.5$

$S_{20}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{20}=\frac{4*(1-1.5^{20})}{1-1.5}$

$S_{20}=26594$

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Example Question #22 : Types Of Series

Calculate the sum, rounded to the nearest integer, of the first 16 terms of the following geometric series: 6+12.6+26.46+116.6886+245.04606...

Possible Answers:

780295

780305

780310

780300

Correct answer:

780305

Explanation:

This is a geometric series.

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

$S_{n} = \frac{a_{1}*(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

We have $a_{1}$ and n and we just need to find r before calculating the sum.

Solution:

$a_{1}=6$

$r=\frac{a_{2}}{a_{1}}$

$r=\frac{12.6}{6} = 2.1$

$S_{16}=\frac{a_{1}*(1-r^n)}{1-r}$

$S_{16}=\frac{6*(1-2.1^{16})}{1-2.1}$

$S_{16}=780305$

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Calculus 2 : Arithmetic and Geometric Series

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2971 : Calculus Ii

Example Question #2972 : Calculus Ii

Example Question #191 : Series In Calculus

Example Question #192 : Series In Calculus

Example Question #2975 : Calculus Ii

Example Question #2981 : Calculus Ii

Example Question #2981 : Calculus Ii

Example Question #21 : Types Of Series

Example Question #2983 : Calculus Ii

Example Question #22 : Types Of Series

All Calculus 2 Resources

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