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Finding the sum of the first 4 terms of a geometric sequence

Problem about sum of arithemtic progression and geometric progressionIs $3^2n$ geometric, if so find common ratio and sum or the first n terms?Finding the common ratio of a geometric series from the sum and first termFinding the first three terms of a geometric sequence, without the first term or common ratio.new sequence formed by adding together corresponding term of a geometric sequence (G.S) and an arithmetic sequence (A.S).The sum of the first $3$ terms is $24$ and the sum of the next $3$ terms is $ 51.$given geometric sequence an arithmetic sequence, find third sequenceFind the total of the sum of the first five terms of the arithmetic series and the sum of the first three terms of the geometric series.Finding arithmetic sequence first termSequences and Series:

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2

1

$begingroup$

The first 3 terms of a geometric sequence are, respectively, the 1st, 4th and 13th terms of an arithmetic sequence.
Given that the first term of the geometric sequence is 3 and the common difference of the arithmetic sequence is 2, calculate the sum of the first 4 terms of the geometric sequence.

What I did was determine the 1st, 4th and 13th terms in the arithmetic sequence to be 3, 9 and 27. I then determined the common ratio for the geo sequence to be 3. This means that the 4th term in the geo sequence is 81. The sum of these 4 terms is 120. I am not 100% confident in my working, so if anyone could confirm whether I have done this correctly, it would be greatly appreciated.

sequences-and-series arithmetic

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asked Jun 15 at 3:35

Justine Justine

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2

1

$begingroup$

The first 3 terms of a geometric sequence are, respectively, the 1st, 4th and 13th terms of an arithmetic sequence.
Given that the first term of the geometric sequence is 3 and the common difference of the arithmetic sequence is 2, calculate the sum of the first 4 terms of the geometric sequence.

What I did was determine the 1st, 4th and 13th terms in the arithmetic sequence to be 3, 9 and 27. I then determined the common ratio for the geo sequence to be 3. This means that the 4th term in the geo sequence is 81. The sum of these 4 terms is 120. I am not 100% confident in my working, so if anyone could confirm whether I have done this correctly, it would be greatly appreciated.

sequences-and-series arithmetic

share|cite|improve this question

asked Jun 15 at 3:35

Justine Justine

3144 bronze badges

New contributor

Justine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

The first 3 terms of a geometric sequence are, respectively, the 1st, 4th and 13th terms of an arithmetic sequence.
Given that the first term of the geometric sequence is 3 and the common difference of the arithmetic sequence is 2, calculate the sum of the first 4 terms of the geometric sequence.

What I did was determine the 1st, 4th and 13th terms in the arithmetic sequence to be 3, 9 and 27. I then determined the common ratio for the geo sequence to be 3. This means that the 4th term in the geo sequence is 81. The sum of these 4 terms is 120. I am not 100% confident in my working, so if anyone could confirm whether I have done this correctly, it would be greatly appreciated.

sequences-and-series arithmetic

share|cite|improve this question

asked Jun 15 at 3:35

Justine Justine

3144 bronze badges

New contributor

Justine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Jun 15 at 3:35

Justine Justine

3144 bronze badges

New contributor

Justine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Jun 15 at 3:35

Justine Justine

3144 bronze badges

New contributor

Justine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Jun 15 at 3:35

Justine Justine

3144 bronze badges

New contributor

Justine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Jun 15 at 3:35

Justine Justine

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2 Answers
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3

$begingroup$

Your reasoning is exactly right. The formula for the $n$th of a geometric sequence is $ar^n$ where $a$ and $r$ are arbitrary constants.

In this case (if we start our numbering at one instead of zero) $a$ is 1 and $r$ is 3. So the terms are the powers of three: 3, 9, 27, 81.

(If you start your numbering at zero, then $a$ is 3 and $r$ is also 3, for the same result. Which version to use depends on your instructor, since they're mathematically equivalent. I find the zero-based one more intuitive, but I'm also a computer scientist more than a mathematician, so take that with a grain of salt.)

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edited Jun 15 at 3:46

answered Jun 15 at 3:41

DraconisDraconis

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3

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$1+x+x^2+x^3=dfrac1-x^41-x$.

So, $1+3+9+27=dfrac 1-811-3=40$.

But your series is $3(1+3+9+27)=3(40)=120$.

share|cite|improve this answer

edited Jun 15 at 3:48

answered Jun 15 at 3:42

Chris CusterChris Custer

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3

$begingroup$

Your reasoning is exactly right. The formula for the $n$th of a geometric sequence is $ar^n$ where $a$ and $r$ are arbitrary constants.

In this case (if we start our numbering at one instead of zero) $a$ is 1 and $r$ is 3. So the terms are the powers of three: 3, 9, 27, 81.

(If you start your numbering at zero, then $a$ is 3 and $r$ is also 3, for the same result. Which version to use depends on your instructor, since they're mathematically equivalent. I find the zero-based one more intuitive, but I'm also a computer scientist more than a mathematician, so take that with a grain of salt.)

share|cite|improve this answer

edited Jun 15 at 3:46

answered Jun 15 at 3:41

DraconisDraconis

54033 silver badges99 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

Your reasoning is exactly right. The formula for the $n$th of a geometric sequence is $ar^n$ where $a$ and $r$ are arbitrary constants.

In this case (if we start our numbering at one instead of zero) $a$ is 1 and $r$ is 3. So the terms are the powers of three: 3, 9, 27, 81.

(If you start your numbering at zero, then $a$ is 3 and $r$ is also 3, for the same result. Which version to use depends on your instructor, since they're mathematically equivalent. I find the zero-based one more intuitive, but I'm also a computer scientist more than a mathematician, so take that with a grain of salt.)

share|cite|improve this answer

edited Jun 15 at 3:46

answered Jun 15 at 3:41

DraconisDraconis

54033 silver badges99 bronze badges

$endgroup$

add a comment | 

$begingroup$

Your reasoning is exactly right. The formula for the $n$th of a geometric sequence is $ar^n$ where $a$ and $r$ are arbitrary constants.

In this case (if we start our numbering at one instead of zero) $a$ is 1 and $r$ is 3. So the terms are the powers of three: 3, 9, 27, 81.

(If you start your numbering at zero, then $a$ is 3 and $r$ is also 3, for the same result. Which version to use depends on your instructor, since they're mathematically equivalent. I find the zero-based one more intuitive, but I'm also a computer scientist more than a mathematician, so take that with a grain of salt.)

share|cite|improve this answer

edited Jun 15 at 3:46

answered Jun 15 at 3:41

DraconisDraconis

54033 silver badges99 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 15 at 3:46

answered Jun 15 at 3:41

DraconisDraconis

54033 silver badges99 bronze badges

answered Jun 15 at 3:41

DraconisDraconis

54033 silver badges99 bronze badges

answered Jun 15 at 3:41

DraconisDraconis

54033 silver badges99 bronze badges

3

$begingroup$

$1+x+x^2+x^3=dfrac1-x^41-x$.

So, $1+3+9+27=dfrac 1-811-3=40$.

But your series is $3(1+3+9+27)=3(40)=120$.

share|cite|improve this answer

edited Jun 15 at 3:48

answered Jun 15 at 3:42

Chris CusterChris Custer

16.8k33 gold badges88 silver badges2929 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

$1+x+x^2+x^3=dfrac1-x^41-x$.

So, $1+3+9+27=dfrac 1-811-3=40$.

But your series is $3(1+3+9+27)=3(40)=120$.

share|cite|improve this answer

edited Jun 15 at 3:48

answered Jun 15 at 3:42

Chris CusterChris Custer

16.8k33 gold badges88 silver badges2929 bronze badges

$endgroup$

add a comment | 

$begingroup$

$1+x+x^2+x^3=dfrac1-x^41-x$.

So, $1+3+9+27=dfrac 1-811-3=40$.

But your series is $3(1+3+9+27)=3(40)=120$.

share|cite|improve this answer

edited Jun 15 at 3:48

answered Jun 15 at 3:42

Chris CusterChris Custer

16.8k33 gold badges88 silver badges2929 bronze badges

$endgroup$

share|cite|improve this answer

edited Jun 15 at 3:48

answered Jun 15 at 3:42

Chris CusterChris Custer

16.8k33 gold badges88 silver badges2929 bronze badges

answered Jun 15 at 3:42

Chris CusterChris Custer

16.8k33 gold badges88 silver badges2929 bronze badges

answered Jun 15 at 3:42

Chris CusterChris Custer

16.8k33 gold badges88 silver badges2929 bronze badges