Divide 56 in four parts in AP such that the ratio of product of their extremes 1st and 4th to product of means 2nd and 3rd is 5:6
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the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
sequences-and-series arithmetic-progressions
asked Aug 20 at 10:27
user2625361
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up vote
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down vote
favorite
the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
sequences-and-series arithmetic-progressions
asked Aug 20 at 10:27
user2625361
62
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it?
– 5xum
Aug 20 at 10:54
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Aug 20 at 10:54
the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
– user2625361
Aug 20 at 11:06
Please, edit your question instead of posting information that is vital to the question in comments.
– 5xum
Aug 20 at 11:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
sequences-and-series arithmetic-progressions
asked Aug 20 at 10:27
user2625361
62
the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
sequences-and-series arithmetic-progressions
asked Aug 20 at 10:27
user2625361
62
edited Aug 20 at 11:12
asked Aug 20 at 10:27
user2625361
62
asked Aug 20 at 10:27
user2625361
62
asked Aug 20 at 10:27
user2625361
62
62
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it?
– 5xum
Aug 20 at 10:54
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Aug 20 at 10:54
the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
– user2625361
Aug 20 at 11:06
Please, edit your question instead of posting information that is vital to the question in comments.
– 5xum
Aug 20 at 11:07
add a comment |Â
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it?
– 5xum
Aug 20 at 10:54
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Aug 20 at 10:54
the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
– user2625361
Aug 20 at 11:06
Please, edit your question instead of posting information that is vital to the question in comments.
– 5xum
Aug 20 at 11:07
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it?
– 5xum
Aug 20 at 10:54
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it?
– 5xum
Aug 20 at 10:54
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Aug 20 at 10:54
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Aug 20 at 10:54
the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
– user2625361
Aug 20 at 11:06
the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
– user2625361
Aug 20 at 11:06
Please, edit your question instead of posting information that is vital to the question in comments.
– 5xum
Aug 20 at 11:07
Please, edit your question instead of posting information that is vital to the question in comments.
– 5xum
Aug 20 at 11:07
add a comment |Â
1 Answer
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You can call the four numbers $a$, $a+d$, $a+2d$, $a+3d$ if you like. This won't affect the final answer although it might make the intermediate algebra more complex.
So we have
$6a(a+3d) = 5(a+d)(a+2d)$
$Rightarrow 6a^2 + 18ad = 5a^2 + 15ad + 10d^2$
$Rightarrow a^2 + 3ad - 10d^2 = 0$
$Rightarrow (a+5d)(a-2d) =0$
So $a=-5d$ or $a=2d$. We also know that the sum of the four terms is 56, so
$4a+6d=56$
From this you can find two alternative pairs of values for $a$ and $d$. In fact, these represent the same four terms, but one sequence is the reverse of the other. This makes sense given the symmetry of the original problem.
answered Aug 20 at 11:39
gandalf61
5,956522
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You can call the four numbers $a$, $a+d$, $a+2d$, $a+3d$ if you like. This won't affect the final answer although it might make the intermediate algebra more complex.
So we have
$6a(a+3d) = 5(a+d)(a+2d)$
$Rightarrow 6a^2 + 18ad = 5a^2 + 15ad + 10d^2$
$Rightarrow a^2 + 3ad - 10d^2 = 0$
$Rightarrow (a+5d)(a-2d) =0$
So $a=-5d$ or $a=2d$. We also know that the sum of the four terms is 56, so
$4a+6d=56$
From this you can find two alternative pairs of values for $a$ and $d$. In fact, these represent the same four terms, but one sequence is the reverse of the other. This makes sense given the symmetry of the original problem.
answered Aug 20 at 11:39
gandalf61
5,956522
add a comment |Â
up vote
0
down vote
You can call the four numbers $a$, $a+d$, $a+2d$, $a+3d$ if you like. This won't affect the final answer although it might make the intermediate algebra more complex.
So we have
$6a(a+3d) = 5(a+d)(a+2d)$
$Rightarrow 6a^2 + 18ad = 5a^2 + 15ad + 10d^2$
$Rightarrow a^2 + 3ad - 10d^2 = 0$
$Rightarrow (a+5d)(a-2d) =0$
So $a=-5d$ or $a=2d$. We also know that the sum of the four terms is 56, so
$4a+6d=56$
From this you can find two alternative pairs of values for $a$ and $d$. In fact, these represent the same four terms, but one sequence is the reverse of the other. This makes sense given the symmetry of the original problem.
answered Aug 20 at 11:39
gandalf61
5,956522
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can call the four numbers $a$, $a+d$, $a+2d$, $a+3d$ if you like. This won't affect the final answer although it might make the intermediate algebra more complex.
So we have
$6a(a+3d) = 5(a+d)(a+2d)$
$Rightarrow 6a^2 + 18ad = 5a^2 + 15ad + 10d^2$
$Rightarrow a^2 + 3ad - 10d^2 = 0$
$Rightarrow (a+5d)(a-2d) =0$
So $a=-5d$ or $a=2d$. We also know that the sum of the four terms is 56, so
$4a+6d=56$
From this you can find two alternative pairs of values for $a$ and $d$. In fact, these represent the same four terms, but one sequence is the reverse of the other. This makes sense given the symmetry of the original problem.
answered Aug 20 at 11:39
gandalf61
5,956522
You can call the four numbers $a$, $a+d$, $a+2d$, $a+3d$ if you like. This won't affect the final answer although it might make the intermediate algebra more complex.
So we have
$6a(a+3d) = 5(a+d)(a+2d)$
$Rightarrow 6a^2 + 18ad = 5a^2 + 15ad + 10d^2$
$Rightarrow a^2 + 3ad - 10d^2 = 0$
$Rightarrow (a+5d)(a-2d) =0$
So $a=-5d$ or $a=2d$. We also know that the sum of the four terms is 56, so
$4a+6d=56$
From this you can find two alternative pairs of values for $a$ and $d$. In fact, these represent the same four terms, but one sequence is the reverse of the other. This makes sense given the symmetry of the original problem.
answered Aug 20 at 11:39
gandalf61
5,956522
answered Aug 20 at 11:39
gandalf61
5,956522
answered Aug 20 at 11:39
gandalf61
5,956522
answered Aug 20 at 11:39
gandalf61
5,956522
5,956522
add a comment |Â
add a comment |Â
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Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it?
– 5xum
Aug 20 at 10:54
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Aug 20 at 10:54
the solution is available on other sources but everyone take the four numbers as (a-3d) , (a-d) , (a+d) , (a+3d). I tried taking other numbers but failed to solve the question. I just need an explanation about the (a-3d) , (a-d) , (a+d) , (a+3d) part , and why cant I get a answer if I take a , a+d , a+2d , a+3d
– user2625361
Aug 20 at 11:06
Please, edit your question instead of posting information that is vital to the question in comments.
– 5xum
Aug 20 at 11:07