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Find the arithmetic sequence a, b, c, d if a−2, b−4, c−3, d+2 is a geometric sequence.
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hi cooljackiec
I'm hoping I've slipped up with my arithmetic because I'm getting a complex number for the common difference. The method should work though so give it a try:
AP
with e = the common difference.
GP
with r = common ratio.
So eliminate b, c and d from the GP equations and then divide the first by the second and first by the third to eliminate r and leave a pair containing just a and e.
I found that it was easiest to then eliminate 'a' to leave a quadratic in e.
But it only had complex solutions so I'll try again carefully.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Several corrections later, I have a solution. The quadratic factorised to give two values of e.
solution 1 gave the required AP and GP.
solution 2 did not. I haven't worked out why not yet.
So I have a single solution. Do you?
LATER EDIT: Looks like solution 2 results from a division by zero and so is not valid.
Bob
Last edited by Bob (2015-03-24 07:51:53)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I used the method before I posted but didn't get anywhere
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Hi Bob
I'm getting just one solution but it has nothing to do with complex numbers.
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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
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hi Stefy
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Yes, that is it.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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hi Stefy,
Thanks. But I'm getting strange results from the second solution to the quadratic. And I've tried two more examples with the same happening again. I'm just trying to make sense of all my paperwork; then I'll post what I've found.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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with r = common ratio.
So eliminate b, c and d from the GP equations and then divide the first by the second and first by the third to eliminate r and leave a pair containing just a and e.
How exactly do you "eliminate b"
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You put in b=a+e.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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I got (a+2e-3)^2=(a+3e+2)(a+e-4) which gives 10a+e^2+22e-4=0
Last edited by cooljackiec (2015-03-25 09:30:44)
I see you have graph paper.
You must be plotting something
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OK. Like this:
AP: a, a+e, a+2e, a+3e
GP: a-2, b-4, c-3, d+2 becomes a-2, a+e-4, a+2e-3, a+3e+2
Now using the common ratio r
The first pair leads to an equation in a and e and the first and third similarly.
Make a the subject of each and equate them to get a quadratic in e. I got
11 leads to the solution you are seeking.
Then my difficulty begins because the second value of e leads to an AP of 2,4,6,8 but not a GP ... values of 0,0,3,10.
So what's going on there.
I made up a second problem: AP 27,31,35,39 and GP 27,36,48,64.
Using the same method I got a quadratic:
e=4 gives the AP and GP as expected but e = -5 gives AP 0,-5,-10,-15 and not a GP of 0,0,3,10 again.
So I picked a third example:
AP 8,13,18,23 and GP 8,12,18,27
My quadratic this time was
e=5 gives the expected AP and GP but e = 1 gives an AP of 0,1,2,3 but not a GP of 0,0,2,7.
I don't understand why this is happening.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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hi Stefy,
Thanks. But I'm getting strange results from the second solution to the quadratic. And I've tried two more examples with the same happening again. I'm just trying to make sense of all my paperwork; then I'll post what I've found.
Bob
I didn't have to solve any quadratics...
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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hi Stefy,
It would help me if you were to post your method. Thanks,
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
Offline