169.If An=200+0.2*A(n-1), and A1=200, A50 will fall in which of the following regions?
(230,239), (240,250), (260,280) .......
oa B
sequence (taught)
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- earth@work
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An=200+0.2*A(n-1)
A50=200+0.2*A49 = 200+0.2*(200+0.2*A48) = 200+40+0.04(A48) = 240+0.04(200+0.2A47) = 240+8+0.008*A47 = 248+0.008*(200+0.2*A46) = 248+1.6+0.0016*A46=249.6+0.0016*A46
This shows that as we carry forward this operation the value will only further increase in decimal value.... so (240,250) is the best option!
maybe there is some other method as i didn't use A1=200 ???
A50=200+0.2*A49 = 200+0.2*(200+0.2*A48) = 200+40+0.04(A48) = 240+0.04(200+0.2A47) = 240+8+0.008*A47 = 248+0.008*(200+0.2*A46) = 248+1.6+0.0016*A46=249.6+0.0016*A46
This shows that as we carry forward this operation the value will only further increase in decimal value.... so (240,250) is the best option!
maybe there is some other method as i didn't use A1=200 ???
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- Ian Stewart
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We have:
a1 = 200
a2 = 240
and the sequence is constantly increasing (we are adding 200 to larger and larger values the further we get into the sequence), so a50 must be greater than 240. In addition, we find the next term (a_N) in the sequence by adding 20% of the previous term (a_(N-1)) to 200. Since we do not yet have a term in our sequence which is 250 or greater, we will never add a number as large as 0.2*250 = 50 to 200 to make a term - there will therefore never be a term as large as 250 in our sequence. The values will, however, get very close to 250. So after a2, every term is somewhere between 240 and 250.
a1 = 200
a2 = 240
and the sequence is constantly increasing (we are adding 200 to larger and larger values the further we get into the sequence), so a50 must be greater than 240. In addition, we find the next term (a_N) in the sequence by adding 20% of the previous term (a_(N-1)) to 200. Since we do not yet have a term in our sequence which is 250 or greater, we will never add a number as large as 0.2*250 = 50 to 200 to make a term - there will therefore never be a term as large as 250 in our sequence. The values will, however, get very close to 250. So after a2, every term is somewhere between 240 and 250.
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An = 200 + A(n-1)/5
A50 = 200 + (40 + 40/5 + 40/5^2 +.....)
A50 = 200 + 40 ( 1 + 1/5 + 1/5^2 + 1/5^3 +........)
A50 = 200 + 40 ( 1-(1/5)^50)/1-1/5
= 200 + 50( 1-(1/5)^50)
Upper limit is 250
Answer is B
A50 = 200 + (40 + 40/5 + 40/5^2 +.....)
A50 = 200 + 40 ( 1 + 1/5 + 1/5^2 + 1/5^3 +........)
A50 = 200 + 40 ( 1-(1/5)^50)/1-1/5
= 200 + 50( 1-(1/5)^50)
Upper limit is 250
Answer is B
- Uri
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Stewart, although I understand that the terms are gradually getting closer to 250, could you please explain the fundamental on which you told that the term will never cross 250?Ian Stewart wrote:Since we do not yet have a term in our sequence which is 250 or greater, we will never add a number as large as 0.2*250 = 50 to 200 to make a term - there will therefore never be a term as large as 250 in our sequence. The values will, however, get very close to 250. So after a2, every term is somewhere between 240 and 250.
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I can try:Uri wrote:Stewart, although I understand that the terms are gradually getting closer to 250, could you please explain the fundamental on which you told that the term will never cross 250?Ian Stewart wrote:Since we do not yet have a term in our sequence which is 250 or greater, we will never add a number as large as 0.2*250 = 50 to 200 to make a term - there will therefore never be a term as large as 250 in our sequence. The values will, however, get very close to 250. So after a2, every term is somewhere between 240 and 250.
Say our sequence is a, b, c, d, e, f, g, ...
We know that we find the next term of our sequence by adding 20% of the previous term to 200. So if we know a, we'd find b by adding 20% of a and 200. Now, if a is less than 250, 20% of a must be less than 50, so b will be less than 250. If b is less than 250, then 20% of b will be less than 50, so c will be less than 250. And so on...
Hope that makes sense - it's probably easier to see if you start with 200 and find the next two or three terms; working out the first few terms is almost always a good idea in sequence questions.
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- Uri
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Ian Stewart wrote:it's probably easier to see if you start with 200 and find the next two or three terms; working out the first few terms is almost always a good idea in sequence questions.
Thanks, Stewart! I solved the problem by finding the first few terms and the answer I guessed was right too! But your logic seemed to be a good one in the long run. Your explanation has made it clear now. Nice effort. Thanks a lot
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