Mixing
find the first term of the series?
$begingroup$
The sum of an infinite geometric series of real numbers is $14,$ and the
sum of the cubes of the terms of this series is $392$.
What is the first term of the series?
My attempt: Let the series be ${ a ,~ ar ,~ar^2 ,ldots }$, then the sum is
$$ s = frac{a}{1 - r} = 14 tag{1}$$
When cubed, the new series is $ a^3,~a^3r^3,~ a^3 r^6, ldots$
which sums to
$$ frac{ a^3 }{ 1 - r^3 } = 392 tag{2}$$
Now I got $frac{a^2}{28}= 1+r+r^2$ after that I'm not able proceed further.
Any hints/solution is appreciated.
sequences-and-series algebra-precalculus summation geometric-progressions
edited Jan 3 at 17:05
Lee David Chung Lin
3,86531140
asked Jan 1 at 21:14
jasminejasmine
1,650416
$endgroup$
add a comment |
$begingroup$
The sum of an infinite geometric series of real numbers is $14,$ and the
sum of the cubes of the terms of this series is $392$.
What is the first term of the series?
My attempt: Let the series be ${ a ,~ ar ,~ar^2 ,ldots }$, then the sum is
$$ s = frac{a}{1 - r} = 14 tag{1}$$
When cubed, the new series is $ a^3,~a^3r^3,~ a^3 r^6, ldots$
which sums to
$$ frac{ a^3 }{ 1 - r^3 } = 392 tag{2}$$
Now I got $frac{a^2}{28}= 1+r+r^2$ after that I'm not able proceed further.
Any hints/solution is appreciated.
sequences-and-series algebra-precalculus summation geometric-progressions
edited Jan 3 at 17:05
Lee David Chung Lin
3,86531140
asked Jan 1 at 21:14
jasminejasmine
1,650416
$endgroup$
add a comment |
$begingroup$
The sum of an infinite geometric series of real numbers is $14,$ and the
sum of the cubes of the terms of this series is $392$.
What is the first term of the series?
My attempt: Let the series be ${ a ,~ ar ,~ar^2 ,ldots }$, then the sum is
$$ s = frac{a}{1 - r} = 14 tag{1}$$
When cubed, the new series is $ a^3,~a^3r^3,~ a^3 r^6, ldots$
which sums to
$$ frac{ a^3 }{ 1 - r^3 } = 392 tag{2}$$
Now I got $frac{a^2}{28}= 1+r+r^2$ after that I'm not able proceed further.
Any hints/solution is appreciated.
sequences-and-series algebra-precalculus summation geometric-progressions
edited Jan 3 at 17:05
Lee David Chung Lin
3,86531140
asked Jan 1 at 21:14
jasminejasmine
1,650416
$endgroup$
The sum of an infinite geometric series of real numbers is $14,$ and the
sum of the cubes of the terms of this series is $392$.
What is the first term of the series?
My attempt: Let the series be ${ a ,~ ar ,~ar^2 ,ldots }$, then the sum is
$$ s = frac{a}{1 - r} = 14 tag{1}$$
When cubed, the new series is $ a^3,~a^3r^3,~ a^3 r^6, ldots$
which sums to
$$ frac{ a^3 }{ 1 - r^3 } = 392 tag{2}$$
Now I got $frac{a^2}{28}= 1+r+r^2$ after that I'm not able proceed further.
Any hints/solution is appreciated.
sequences-and-series algebra-precalculus summation geometric-progressions
sequences-and-series algebra-precalculus summation geometric-progressions
edited Jan 3 at 17:05
Lee David Chung Lin
3,86531140
asked Jan 1 at 21:14
jasminejasmine
1,650416
edited Jan 3 at 17:05
Lee David Chung Lin
3,86531140
asked Jan 1 at 21:14
jasminejasmine
1,650416
edited Jan 3 at 17:05
Lee David Chung Lin
3,86531140
edited Jan 3 at 17:05
Lee David Chung Lin
3,86531140
edited Jan 3 at 17:05
Lee David Chung Lin
3,86531140
3,86531140
asked Jan 1 at 21:14
jasminejasmine
1,650416
asked Jan 1 at 21:14
jasminejasmine
1,650416
asked Jan 1 at 21:14
jasminejasmine
1,650416
1,650416
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
$$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
and so
$$(2r-1)(r-2)=0implies r=1/2$$
since $|r|<1.$ And so $a=7.$
answered Jan 1 at 21:21
Hello_WorldHello_World
4,16221631
$endgroup$
$begingroup$
how $1+ r+r^2 = 1 +r^2-2r$? im not getting
$endgroup$
– jasmine
Jan 1 at 21:26
1
$begingroup$
They are not equal. $28(1+r+r^2) = 196(1-r)^2$
$endgroup$
– Hello_World
Jan 1 at 21:30
$begingroup$
okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
$endgroup$
– jasmine
Jan 1 at 21:38
add a comment |
$begingroup$
You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.
answered Jan 1 at 21:23
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058889%2ffind-the-first-term-of-the-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
$$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
and so
$$(2r-1)(r-2)=0implies r=1/2$$
since $|r|<1.$ And so $a=7.$
answered Jan 1 at 21:21
Hello_WorldHello_World
4,16221631
$endgroup$
$begingroup$
how $1+ r+r^2 = 1 +r^2-2r$? im not getting
$endgroup$
– jasmine
Jan 1 at 21:26
1
$begingroup$
They are not equal. $28(1+r+r^2) = 196(1-r)^2$
$endgroup$
– Hello_World
Jan 1 at 21:30
$begingroup$
okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
$endgroup$
– jasmine
Jan 1 at 21:38
add a comment |
$begingroup$
So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
$$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
and so
$$(2r-1)(r-2)=0implies r=1/2$$
since $|r|<1.$ And so $a=7.$
answered Jan 1 at 21:21
Hello_WorldHello_World
4,16221631
$endgroup$
$begingroup$
how $1+ r+r^2 = 1 +r^2-2r$? im not getting
$endgroup$
– jasmine
Jan 1 at 21:26
1
$begingroup$
They are not equal. $28(1+r+r^2) = 196(1-r)^2$
$endgroup$
– Hello_World
Jan 1 at 21:30
$begingroup$
okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
$endgroup$
– jasmine
Jan 1 at 21:38
add a comment |
$begingroup$
So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
$$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
and so
$$(2r-1)(r-2)=0implies r=1/2$$
since $|r|<1.$ And so $a=7.$
answered Jan 1 at 21:21
Hello_WorldHello_World
4,16221631
$endgroup$
So you have $a=14(1-r)$ and $a^3 = (1-r)(1+r+r^2)392$ which implies
$$a^2=28(1+r+r^2)=196(1+r^2-2r)$$
and so
$$(2r-1)(r-2)=0implies r=1/2$$
since $|r|<1.$ And so $a=7.$
answered Jan 1 at 21:21
Hello_WorldHello_World
4,16221631
answered Jan 1 at 21:21
Hello_WorldHello_World
4,16221631
answered Jan 1 at 21:21
Hello_WorldHello_World
4,16221631
answered Jan 1 at 21:21
Hello_WorldHello_World
4,16221631
4,16221631
$begingroup$
how $1+ r+r^2 = 1 +r^2-2r$? im not getting
$endgroup$
– jasmine
Jan 1 at 21:26
1
$begingroup$
They are not equal. $28(1+r+r^2) = 196(1-r)^2$
$endgroup$
– Hello_World
Jan 1 at 21:30
$begingroup$
okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
$endgroup$
– jasmine
Jan 1 at 21:38
add a comment |
$begingroup$
how $1+ r+r^2 = 1 +r^2-2r$? im not getting
$endgroup$
– jasmine
Jan 1 at 21:26
1
$begingroup$
They are not equal. $28(1+r+r^2) = 196(1-r)^2$
$endgroup$
– Hello_World
Jan 1 at 21:30
$begingroup$
okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
$endgroup$
– jasmine
Jan 1 at 21:38
$begingroup$
how $1+ r+r^2 = 1 +r^2-2r$? im not getting
$endgroup$
– jasmine
Jan 1 at 21:26
$begingroup$
how $1+ r+r^2 = 1 +r^2-2r$? im not getting
$endgroup$
– jasmine
Jan 1 at 21:26
1
1
$begingroup$
They are not equal. $28(1+r+r^2) = 196(1-r)^2$
$endgroup$
– Hello_World
Jan 1 at 21:30
$begingroup$
They are not equal. $28(1+r+r^2) = 196(1-r)^2$
$endgroup$
– Hello_World
Jan 1 at 21:30
$begingroup$
okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
$endgroup$
– jasmine
Jan 1 at 21:38
$begingroup$
okkss .....my eyes didn't see 196,,,as i read so fast ur answer ....and at same moment i was looking jose carlos sir answer
$endgroup$
– jasmine
Jan 1 at 21:38
add a comment |
$begingroup$
You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.
answered Jan 1 at 21:23
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
add a comment |
$begingroup$
You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.
answered Jan 1 at 21:23
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
add a comment |
$begingroup$
You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.
answered Jan 1 at 21:23
José Carlos SantosJosé Carlos Santos
153k22123225
$endgroup$
You started fine. This leads you to the system$$left{begin{array}{l}dfrac a{1-r}=14\dfrac{a^3}{1-r^3}=392.end{array}right.$$This system has two solutions: $(a,r)=(-14,2)$ and $(a,r)=left(7,frac12right)$. But only the second one leads to convergent series.
answered Jan 1 at 21:23
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jan 1 at 21:23
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jan 1 at 21:23
José Carlos SantosJosé Carlos Santos
153k22123225
answered Jan 1 at 21:23
José Carlos SantosJosé Carlos Santos
153k22123225
153k22123225
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3058889%2ffind-the-first-term-of-the-series%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
