Problem on AP/GP
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Could you please define Arithmetic Progression - mathematically?
Could you please define Geometric Progression - mathematically?
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Arithmetic progression is a series of terms where the difference between the terms are constant. Geometric progression is a series of terms where the ratio between the terms are constant. |
D
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Arithmetic progression is a series of terms where the difference between the terms are constant. - Correct
Please write an example of AP, using 'a' as the first term and 'd' as the 'difference'. Write four terms of the series.
Geometric progression is a series of terms where the ratio between the terms are constant. - Correct
Please write an example of AP, using 'a1' as the first term and 'r' as the the 'ratio' between the terms. Write four terms of the series.
This problem seems intentionally worded deceptively: "2x, and 2(x - y), x [MATH]\ne[/MATH] 0, are respectively the first three terms of a geometric series" provides only two terms.
Are they terms 1 and 2 or terms 2 and 3. If x is the first term, the common ration would be 2 rather than 0.5.
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It is unnecessarily deceptive! Since we can calculate 'y' from the AP - we do not need the third term for GP. It should not have "spoken" of the third term.
Steven G
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I am having trouble with this. What am I missing? The difference between x and (x+y) is y. So the common difference is y. The third term is 2x+2 = x+(x+2) = x+2y, so 2y = x+2.
GP: 2x, 2(x-y) = 2x-2y = 2x-(x+2) = x-2. But why is (x-2)/(2x) = .5??
GP: 2x, 2(x-y) = 2x-2y = 2x-(x+2) = x-2. But why is (x-2)/(2x) = .5??
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It does not matter if we have three terms. Both Arithmetic and Geometric progressions require only two terms to pin it down. Sure, the problem statement seems flawed. This does not necessarily prevent a rational solution. Maybe it does, but see what there is to see. Don't just throw the problem to the dogs!
For the Geometric
1) 2x
2) 2(x-y)
2x * What = 2(x-y) ==> What = 2(x-y)/2x = (x-y)/x <== This is why we need x != 0.
Is (x-y)/x = 0.5? We may be stuck at this point, but we made a rational effort with what we had.
If we pick x = 1 and y = 1/2, what do we have? I don't see any requirements for x or y, so why not just pick some values? Just don't pick x = 0.
Want extra credit on an exam? Come up with some sort of decent solution on a flawed problem statement. That should do it!
For the Geometric
1) 2x
2) 2(x-y)
2x * What = 2(x-y) ==> What = 2(x-y)/2x = (x-y)/x <== This is why we need x != 0.
Is (x-y)/x = 0.5? We may be stuck at this point, but we made a rational effort with what we had.
If we pick x = 1 and y = 1/2, what do we have? I don't see any requirements for x or y, so why not just pick some values? Just don't pick x = 0.
Want extra credit on an exam? Come up with some sort of decent solution on a flawed problem statement. That should do it!
Steven G
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Yes x≠0 is a condition. What do you think is going on? Can you find the common difference in terms of x?
pka
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I too think that the second part (the geometric series) is flawed.
The arithmetic series is \(\displaystyle a_n=x+(n-1)(0.5x+1)\)
e) the sum of the first 121 terms is \(\displaystyle \sum\limits_{k = 0}^{120} {\left\{ {x + k\left( {0.5x + 1} \right)} \right\}} \) SEE HERE
f) The \(\displaystyle 87^{th}\) term is \(\displaystyle x+86(0.5x+1)\)
SEE HERE
The arithmetic series is \(\displaystyle a_n=x+(n-1)(0.5x+1)\)
e) the sum of the first 121 terms is \(\displaystyle \sum\limits_{k = 0}^{120} {\left\{ {x + k\left( {0.5x + 1} \right)} \right\}} \) SEE HERE
f) The \(\displaystyle 87^{th}\) term is \(\displaystyle x+86(0.5x+1)\)
SEE HERE
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