Find the common ratio, if the sum of the first $8$ terms in a geometric progression is equal to $17$ times the sum of its first $4$ terms
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Find the common ratio, if the sum of the first $8$ terms in a geometric progression (GP) is equal to $17$ times the sum of its first $4$ terms.
So far I have got
$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3).$$
How do I proceed without using the sum of GP formula? Thank you.
sequences-and-series geometric-progressions
edited Aug 26 at 12:18
Jendrik Stelzner
7,58221037
asked Aug 26 at 11:44
MERcurialKG
65
add a comment |Â
up vote
0
down vote
favorite
Find the common ratio, if the sum of the first $8$ terms in a geometric progression (GP) is equal to $17$ times the sum of its first $4$ terms.
So far I have got
$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3).$$
How do I proceed without using the sum of GP formula? Thank you.
sequences-and-series geometric-progressions
edited Aug 26 at 12:18
Jendrik Stelzner
7,58221037
asked Aug 26 at 11:44
MERcurialKG
65
2
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 26 at 11:46
Can you correct your typos please?
– Dr. Sonnhard Graubner
Aug 26 at 11:46
I think you meant "of the sum.."
– Dr. Sonnhard Graubner
Aug 26 at 11:47
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the common ratio, if the sum of the first $8$ terms in a geometric progression (GP) is equal to $17$ times the sum of its first $4$ terms.
So far I have got
$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3).$$
How do I proceed without using the sum of GP formula? Thank you.
sequences-and-series geometric-progressions
edited Aug 26 at 12:18
Jendrik Stelzner
7,58221037
asked Aug 26 at 11:44
MERcurialKG
65
Find the common ratio, if the sum of the first $8$ terms in a geometric progression (GP) is equal to $17$ times the sum of its first $4$ terms.
So far I have got
$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3).$$
How do I proceed without using the sum of GP formula? Thank you.
sequences-and-series geometric-progressions
edited Aug 26 at 12:18
Jendrik Stelzner
7,58221037
asked Aug 26 at 11:44
MERcurialKG
65
edited Aug 26 at 12:18
Jendrik Stelzner
7,58221037
edited Aug 26 at 12:18
Jendrik Stelzner
7,58221037
edited Aug 26 at 12:18
Jendrik Stelzner
7,58221037
7,58221037
asked Aug 26 at 11:44
MERcurialKG
65
asked Aug 26 at 11:44
MERcurialKG
65
asked Aug 26 at 11:44
MERcurialKG
65
65
2
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 26 at 11:46
Can you correct your typos please?
– Dr. Sonnhard Graubner
Aug 26 at 11:46
I think you meant "of the sum.."
– Dr. Sonnhard Graubner
Aug 26 at 11:47
add a comment |Â
2
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 26 at 11:46
Can you correct your typos please?
– Dr. Sonnhard Graubner
Aug 26 at 11:46
I think you meant "of the sum.."
– Dr. Sonnhard Graubner
Aug 26 at 11:47
2
2
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 26 at 11:46
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 26 at 11:46
Can you correct your typos please?
– Dr. Sonnhard Graubner
Aug 26 at 11:46
Can you correct your typos please?
– Dr. Sonnhard Graubner
Aug 26 at 11:46
I think you meant "of the sum.."
– Dr. Sonnhard Graubner
Aug 26 at 11:47
I think you meant "of the sum.."
– Dr. Sonnhard Graubner
Aug 26 at 11:47
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Hint:
$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$$
$$a(1+r+r^2+r^3)+ar^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$$
Hint 2:
After the above step, we can gather everything onto the left-hand side and then factor:
$$ar^4(1+r+r^2+r^3)-16a(1+r+r^2+r^3)=0$$
$$a(r^4-16)(1+r+r^2+r^3)=0$$
Can you see what to do from here?
answered Aug 26 at 11:58
Théophile
17k12438
1
Thank you very much, it sure helped!
– MERcurialKG
Aug 27 at 7:16
do you minus a(1+r+r2+r3) from both sides?? wouldnt a(1+r+r2+r3) get cancel out leaving 16a instead?
– MERcurialKG
Aug 30 at 11:08
@MERcurialKG What I did was to subtract $17a(1+r+r^2+r^3)$ from both sides to make the right-hand side $0$. This doesn't cancel out the $(1+r+r^2+r^3)$, though! If it helps, you could write $x = a(1+r+r^2+r^3)$, so that we have $$x+r^4x=17x.$$ Then subtract $17x$ from both sides to get $r^4x-16x=0$. Now, $x$ is still $a(1+r+r^2+r^3)$, so it's important to see that the $(1+r+r^2+r^3)$ is still there. In fact, it would be better to write it out again: using $x$ is only temporarily helpful to get a better grasp of factoring, but we probably shouldn't use it...
– Théophile
Aug 30 at 12:01
... because it unnecessarily introduces a new variable and hides important information about $a$ and $r$. So the moral is, you're allowed to do a trick like this if it helps with factoring, but just remember to change the $x$ back sooner rather than later. For the problem at hand, see if you can factor $1+r+r^2+r^3$ in a similar way to how we did $a+ar+cdots+ar^7$.
– Théophile
Aug 30 at 12:08
add a comment |Â
up vote
-1
down vote
$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$
$a(1+r+r^2+r^3)+a^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$
Cancel out $a(1+r+r^2+r^3)$ to get:
$1+a^3=17$
or, $a^3=16$
Thus, $a=2sqrt[3]2$
answered Aug 26 at 16:13
Chrys
24
Wow thanks! Cannot believe it was that easy!
– MERcurialKG
Aug 27 at 7:15
should i round the answer?
– MERcurialKG
Aug 27 at 7:21
This is incorrect: you can only 'cancel out' a term from each side if it is nonzero. For example, if we had $x^2=x$, then cancelling out an $x$ from each side gives $x=1$, yet we should also have $x=0$. This is why it is safer to factor.
– Théophile
Aug 28 at 1:37
Thanks, @Théophile, also can you please clear things up for me? In chrys answer, he said a4(1+r+r2+r3) is equal to ar4+ar5+ar6+ar7 so if a4 x r3 is equal to ar7??? Sorry and thanks!!!
– MERcurialKG
Aug 29 at 11:32
@MERcurialKG Yes, that is another mistake: certainly $a^4$ was intended to be $ar^4$. I'll add some detail to my own answer to show the steps.
– Théophile
Aug 29 at 14:34
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint:
$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$$
$$a(1+r+r^2+r^3)+ar^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$$
Hint 2:
After the above step, we can gather everything onto the left-hand side and then factor:
$$ar^4(1+r+r^2+r^3)-16a(1+r+r^2+r^3)=0$$
$$a(r^4-16)(1+r+r^2+r^3)=0$$
Can you see what to do from here?
answered Aug 26 at 11:58
Théophile
17k12438
1
Thank you very much, it sure helped!
– MERcurialKG
Aug 27 at 7:16
do you minus a(1+r+r2+r3) from both sides?? wouldnt a(1+r+r2+r3) get cancel out leaving 16a instead?
– MERcurialKG
Aug 30 at 11:08
@MERcurialKG What I did was to subtract $17a(1+r+r^2+r^3)$ from both sides to make the right-hand side $0$. This doesn't cancel out the $(1+r+r^2+r^3)$, though! If it helps, you could write $x = a(1+r+r^2+r^3)$, so that we have $$x+r^4x=17x.$$ Then subtract $17x$ from both sides to get $r^4x-16x=0$. Now, $x$ is still $a(1+r+r^2+r^3)$, so it's important to see that the $(1+r+r^2+r^3)$ is still there. In fact, it would be better to write it out again: using $x$ is only temporarily helpful to get a better grasp of factoring, but we probably shouldn't use it...
– Théophile
Aug 30 at 12:01
... because it unnecessarily introduces a new variable and hides important information about $a$ and $r$. So the moral is, you're allowed to do a trick like this if it helps with factoring, but just remember to change the $x$ back sooner rather than later. For the problem at hand, see if you can factor $1+r+r^2+r^3$ in a similar way to how we did $a+ar+cdots+ar^7$.
– Théophile
Aug 30 at 12:08
add a comment |Â
up vote
2
down vote
Hint:
$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$$
$$a(1+r+r^2+r^3)+ar^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$$
Hint 2:
After the above step, we can gather everything onto the left-hand side and then factor:
$$ar^4(1+r+r^2+r^3)-16a(1+r+r^2+r^3)=0$$
$$a(r^4-16)(1+r+r^2+r^3)=0$$
Can you see what to do from here?
answered Aug 26 at 11:58
Théophile
17k12438
1
Thank you very much, it sure helped!
– MERcurialKG
Aug 27 at 7:16
do you minus a(1+r+r2+r3) from both sides?? wouldnt a(1+r+r2+r3) get cancel out leaving 16a instead?
– MERcurialKG
Aug 30 at 11:08
@MERcurialKG What I did was to subtract $17a(1+r+r^2+r^3)$ from both sides to make the right-hand side $0$. This doesn't cancel out the $(1+r+r^2+r^3)$, though! If it helps, you could write $x = a(1+r+r^2+r^3)$, so that we have $$x+r^4x=17x.$$ Then subtract $17x$ from both sides to get $r^4x-16x=0$. Now, $x$ is still $a(1+r+r^2+r^3)$, so it's important to see that the $(1+r+r^2+r^3)$ is still there. In fact, it would be better to write it out again: using $x$ is only temporarily helpful to get a better grasp of factoring, but we probably shouldn't use it...
– Théophile
Aug 30 at 12:01
... because it unnecessarily introduces a new variable and hides important information about $a$ and $r$. So the moral is, you're allowed to do a trick like this if it helps with factoring, but just remember to change the $x$ back sooner rather than later. For the problem at hand, see if you can factor $1+r+r^2+r^3$ in a similar way to how we did $a+ar+cdots+ar^7$.
– Théophile
Aug 30 at 12:08
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint:
$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$$
$$a(1+r+r^2+r^3)+ar^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$$
Hint 2:
After the above step, we can gather everything onto the left-hand side and then factor:
$$ar^4(1+r+r^2+r^3)-16a(1+r+r^2+r^3)=0$$
$$a(r^4-16)(1+r+r^2+r^3)=0$$
Can you see what to do from here?
answered Aug 26 at 11:58
Théophile
17k12438
Hint:
$$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$$
$$a(1+r+r^2+r^3)+ar^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$$
Hint 2:
After the above step, we can gather everything onto the left-hand side and then factor:
$$ar^4(1+r+r^2+r^3)-16a(1+r+r^2+r^3)=0$$
$$a(r^4-16)(1+r+r^2+r^3)=0$$
Can you see what to do from here?
answered Aug 26 at 11:58
Théophile
17k12438
edited Aug 29 at 14:39
answered Aug 26 at 11:58
Théophile
17k12438
answered Aug 26 at 11:58
Théophile
17k12438
answered Aug 26 at 11:58
Théophile
17k12438
17k12438
1
Thank you very much, it sure helped!
– MERcurialKG
Aug 27 at 7:16
do you minus a(1+r+r2+r3) from both sides?? wouldnt a(1+r+r2+r3) get cancel out leaving 16a instead?
– MERcurialKG
Aug 30 at 11:08
@MERcurialKG What I did was to subtract $17a(1+r+r^2+r^3)$ from both sides to make the right-hand side $0$. This doesn't cancel out the $(1+r+r^2+r^3)$, though! If it helps, you could write $x = a(1+r+r^2+r^3)$, so that we have $$x+r^4x=17x.$$ Then subtract $17x$ from both sides to get $r^4x-16x=0$. Now, $x$ is still $a(1+r+r^2+r^3)$, so it's important to see that the $(1+r+r^2+r^3)$ is still there. In fact, it would be better to write it out again: using $x$ is only temporarily helpful to get a better grasp of factoring, but we probably shouldn't use it...
– Théophile
Aug 30 at 12:01
... because it unnecessarily introduces a new variable and hides important information about $a$ and $r$. So the moral is, you're allowed to do a trick like this if it helps with factoring, but just remember to change the $x$ back sooner rather than later. For the problem at hand, see if you can factor $1+r+r^2+r^3$ in a similar way to how we did $a+ar+cdots+ar^7$.
– Théophile
Aug 30 at 12:08
add a comment |Â
1
Thank you very much, it sure helped!
– MERcurialKG
Aug 27 at 7:16
do you minus a(1+r+r2+r3) from both sides?? wouldnt a(1+r+r2+r3) get cancel out leaving 16a instead?
– MERcurialKG
Aug 30 at 11:08
@MERcurialKG What I did was to subtract $17a(1+r+r^2+r^3)$ from both sides to make the right-hand side $0$. This doesn't cancel out the $(1+r+r^2+r^3)$, though! If it helps, you could write $x = a(1+r+r^2+r^3)$, so that we have $$x+r^4x=17x.$$ Then subtract $17x$ from both sides to get $r^4x-16x=0$. Now, $x$ is still $a(1+r+r^2+r^3)$, so it's important to see that the $(1+r+r^2+r^3)$ is still there. In fact, it would be better to write it out again: using $x$ is only temporarily helpful to get a better grasp of factoring, but we probably shouldn't use it...
– Théophile
Aug 30 at 12:01
... because it unnecessarily introduces a new variable and hides important information about $a$ and $r$. So the moral is, you're allowed to do a trick like this if it helps with factoring, but just remember to change the $x$ back sooner rather than later. For the problem at hand, see if you can factor $1+r+r^2+r^3$ in a similar way to how we did $a+ar+cdots+ar^7$.
– Théophile
Aug 30 at 12:08
1
1
Thank you very much, it sure helped!
– MERcurialKG
Aug 27 at 7:16
Thank you very much, it sure helped!
– MERcurialKG
Aug 27 at 7:16
do you minus a(1+r+r2+r3) from both sides?? wouldnt a(1+r+r2+r3) get cancel out leaving 16a instead?
– MERcurialKG
Aug 30 at 11:08
do you minus a(1+r+r2+r3) from both sides?? wouldnt a(1+r+r2+r3) get cancel out leaving 16a instead?
– MERcurialKG
Aug 30 at 11:08
@MERcurialKG What I did was to subtract $17a(1+r+r^2+r^3)$ from both sides to make the right-hand side $0$. This doesn't cancel out the $(1+r+r^2+r^3)$, though! If it helps, you could write $x = a(1+r+r^2+r^3)$, so that we have $$x+r^4x=17x.$$ Then subtract $17x$ from both sides to get $r^4x-16x=0$. Now, $x$ is still $a(1+r+r^2+r^3)$, so it's important to see that the $(1+r+r^2+r^3)$ is still there. In fact, it would be better to write it out again: using $x$ is only temporarily helpful to get a better grasp of factoring, but we probably shouldn't use it...
– Théophile
Aug 30 at 12:01
@MERcurialKG What I did was to subtract $17a(1+r+r^2+r^3)$ from both sides to make the right-hand side $0$. This doesn't cancel out the $(1+r+r^2+r^3)$, though! If it helps, you could write $x = a(1+r+r^2+r^3)$, so that we have $$x+r^4x=17x.$$ Then subtract $17x$ from both sides to get $r^4x-16x=0$. Now, $x$ is still $a(1+r+r^2+r^3)$, so it's important to see that the $(1+r+r^2+r^3)$ is still there. In fact, it would be better to write it out again: using $x$ is only temporarily helpful to get a better grasp of factoring, but we probably shouldn't use it...
– Théophile
Aug 30 at 12:01
... because it unnecessarily introduces a new variable and hides important information about $a$ and $r$. So the moral is, you're allowed to do a trick like this if it helps with factoring, but just remember to change the $x$ back sooner rather than later. For the problem at hand, see if you can factor $1+r+r^2+r^3$ in a similar way to how we did $a+ar+cdots+ar^7$.
– Théophile
Aug 30 at 12:08
... because it unnecessarily introduces a new variable and hides important information about $a$ and $r$. So the moral is, you're allowed to do a trick like this if it helps with factoring, but just remember to change the $x$ back sooner rather than later. For the problem at hand, see if you can factor $1+r+r^2+r^3$ in a similar way to how we did $a+ar+cdots+ar^7$.
– Théophile
Aug 30 at 12:08
add a comment |Â
up vote
-1
down vote
$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$
$a(1+r+r^2+r^3)+a^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$
Cancel out $a(1+r+r^2+r^3)$ to get:
$1+a^3=17$
or, $a^3=16$
Thus, $a=2sqrt[3]2$
answered Aug 26 at 16:13
Chrys
24
Wow thanks! Cannot believe it was that easy!
– MERcurialKG
Aug 27 at 7:15
should i round the answer?
– MERcurialKG
Aug 27 at 7:21
This is incorrect: you can only 'cancel out' a term from each side if it is nonzero. For example, if we had $x^2=x$, then cancelling out an $x$ from each side gives $x=1$, yet we should also have $x=0$. This is why it is safer to factor.
– Théophile
Aug 28 at 1:37
Thanks, @Théophile, also can you please clear things up for me? In chrys answer, he said a4(1+r+r2+r3) is equal to ar4+ar5+ar6+ar7 so if a4 x r3 is equal to ar7??? Sorry and thanks!!!
– MERcurialKG
Aug 29 at 11:32
@MERcurialKG Yes, that is another mistake: certainly $a^4$ was intended to be $ar^4$. I'll add some detail to my own answer to show the steps.
– Théophile
Aug 29 at 14:34
 |Â
show 1 more comment
up vote
-1
down vote
$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$
$a(1+r+r^2+r^3)+a^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$
Cancel out $a(1+r+r^2+r^3)$ to get:
$1+a^3=17$
or, $a^3=16$
Thus, $a=2sqrt[3]2$
answered Aug 26 at 16:13
Chrys
24
Wow thanks! Cannot believe it was that easy!
– MERcurialKG
Aug 27 at 7:15
should i round the answer?
– MERcurialKG
Aug 27 at 7:21
This is incorrect: you can only 'cancel out' a term from each side if it is nonzero. For example, if we had $x^2=x$, then cancelling out an $x$ from each side gives $x=1$, yet we should also have $x=0$. This is why it is safer to factor.
– Théophile
Aug 28 at 1:37
Thanks, @Théophile, also can you please clear things up for me? In chrys answer, he said a4(1+r+r2+r3) is equal to ar4+ar5+ar6+ar7 so if a4 x r3 is equal to ar7??? Sorry and thanks!!!
– MERcurialKG
Aug 29 at 11:32
@MERcurialKG Yes, that is another mistake: certainly $a^4$ was intended to be $ar^4$. I'll add some detail to my own answer to show the steps.
– Théophile
Aug 29 at 14:34
 |Â
show 1 more comment
up vote
-1
down vote
up vote
-1
down vote
$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$
$a(1+r+r^2+r^3)+a^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$
Cancel out $a(1+r+r^2+r^3)$ to get:
$1+a^3=17$
or, $a^3=16$
Thus, $a=2sqrt[3]2$
answered Aug 26 at 16:13
Chrys
24
$a+ar+ar^2+ar^3+ar^4+ar^5+ar^6+ar^7=17(a+ar+ar^2+ar^3)$
$a(1+r+r^2+r^3)+a^4(1+r+r^2+r^3)=17a(1+r+r^2+r^3)$
Cancel out $a(1+r+r^2+r^3)$ to get:
$1+a^3=17$
or, $a^3=16$
Thus, $a=2sqrt[3]2$
answered Aug 26 at 16:13
Chrys
24
answered Aug 26 at 16:13
Chrys
24
answered Aug 26 at 16:13
Chrys
24
answered Aug 26 at 16:13
Chrys
24
24
Wow thanks! Cannot believe it was that easy!
– MERcurialKG
Aug 27 at 7:15
should i round the answer?
– MERcurialKG
Aug 27 at 7:21
This is incorrect: you can only 'cancel out' a term from each side if it is nonzero. For example, if we had $x^2=x$, then cancelling out an $x$ from each side gives $x=1$, yet we should also have $x=0$. This is why it is safer to factor.
– Théophile
Aug 28 at 1:37
Thanks, @Théophile, also can you please clear things up for me? In chrys answer, he said a4(1+r+r2+r3) is equal to ar4+ar5+ar6+ar7 so if a4 x r3 is equal to ar7??? Sorry and thanks!!!
– MERcurialKG
Aug 29 at 11:32
@MERcurialKG Yes, that is another mistake: certainly $a^4$ was intended to be $ar^4$. I'll add some detail to my own answer to show the steps.
– Théophile
Aug 29 at 14:34
 |Â
show 1 more comment
Wow thanks! Cannot believe it was that easy!
– MERcurialKG
Aug 27 at 7:15
should i round the answer?
– MERcurialKG
Aug 27 at 7:21
This is incorrect: you can only 'cancel out' a term from each side if it is nonzero. For example, if we had $x^2=x$, then cancelling out an $x$ from each side gives $x=1$, yet we should also have $x=0$. This is why it is safer to factor.
– Théophile
Aug 28 at 1:37
Thanks, @Théophile, also can you please clear things up for me? In chrys answer, he said a4(1+r+r2+r3) is equal to ar4+ar5+ar6+ar7 so if a4 x r3 is equal to ar7??? Sorry and thanks!!!
– MERcurialKG
Aug 29 at 11:32
@MERcurialKG Yes, that is another mistake: certainly $a^4$ was intended to be $ar^4$. I'll add some detail to my own answer to show the steps.
– Théophile
Aug 29 at 14:34
Wow thanks! Cannot believe it was that easy!
– MERcurialKG
Aug 27 at 7:15
Wow thanks! Cannot believe it was that easy!
– MERcurialKG
Aug 27 at 7:15
should i round the answer?
– MERcurialKG
Aug 27 at 7:21
should i round the answer?
– MERcurialKG
Aug 27 at 7:21
This is incorrect: you can only 'cancel out' a term from each side if it is nonzero. For example, if we had $x^2=x$, then cancelling out an $x$ from each side gives $x=1$, yet we should also have $x=0$. This is why it is safer to factor.
– Théophile
Aug 28 at 1:37
This is incorrect: you can only 'cancel out' a term from each side if it is nonzero. For example, if we had $x^2=x$, then cancelling out an $x$ from each side gives $x=1$, yet we should also have $x=0$. This is why it is safer to factor.
– Théophile
Aug 28 at 1:37
Thanks, @Théophile, also can you please clear things up for me? In chrys answer, he said a4(1+r+r2+r3) is equal to ar4+ar5+ar6+ar7 so if a4 x r3 is equal to ar7??? Sorry and thanks!!!
– MERcurialKG
Aug 29 at 11:32
Thanks, @Théophile, also can you please clear things up for me? In chrys answer, he said a4(1+r+r2+r3) is equal to ar4+ar5+ar6+ar7 so if a4 x r3 is equal to ar7??? Sorry and thanks!!!
– MERcurialKG
Aug 29 at 11:32
@MERcurialKG Yes, that is another mistake: certainly $a^4$ was intended to be $ar^4$. I'll add some detail to my own answer to show the steps.
– Théophile
Aug 29 at 14:34
@MERcurialKG Yes, that is another mistake: certainly $a^4$ was intended to be $ar^4$. I'll add some detail to my own answer to show the steps.
– Théophile
Aug 29 at 14:34
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Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 26 at 11:46
Can you correct your typos please?
– Dr. Sonnhard Graubner
Aug 26 at 11:46
I think you meant "of the sum.."
– Dr. Sonnhard Graubner
Aug 26 at 11:47