Number series: increase by percentage

[hiyo]

I don't know if this question belongs in this section, so sorry if it doesn't. This is about a number series.


I need to be able to take a number (R) and spread it out over a number of points (X) such that the final value is a multiple (Y) of the first value, and find the percentage change (Z) from value to value to make this possible.


Example: I want to spread the number 1000 out over 100 points. The last point needs to be 3.3x the first point. What is the percentage increase from point to point to create this distribution?


Any tips about software that can do this easily would be most appreciated. I'd love to have all the values displayed too so I can input them without stopping to calculate. This is for my job.


Thanks!

 

[Country Boy]

You don't say that the values rise by the same amount from point to point but without that there is not enough information. If we call the initial value "a" and the increase from point to point "r" then the values are a, a+ r, a+ 2r, ..., a+ 99r. The fact that the final amount, a+ 99r, is 3.3 times the initial amount, a, means that a+ 99r= 3.3a.

Subtracting a from both sides, 99r= 2.3a.

We also have the fact that we are "spreading out 1000" which I take to mean that these numbers add to 1000: a+ (a+ r)+ (a+ 2r)+ ...+ a+ 99r= 100a+ (1+ 2+ ...+ 99)r.= 1000.

To sum that last arithmetic series write
S= 1+ 2+ 3+ ...+ 97+ 98+ 99 and reverse the order of the numbers
S= 99+ 98+ 97+ ... + 3+ 2+ 1 . Now sum vertically.
2S= 100+ 100+ 100+ ...+ 100+ 100+ 100= 99(100)= 9900.
So S= 4950 and we have 100a+ 4950r= 1000.

Solve 99r= 2.3a and 100a+ 4950r= 1000.
r= (2.3/99)a= 0.023a so 100a+ 4950r= 100a+ 115a= 215a= 1000 and a= 1000/215= 8.69 approximately.

Then r= (2.3/99)(8.69)= 0.2020.

 

[nycmathguy]

You don't say that the values rise by the same amount from point to point but without that there is not enough information. If we call the initial value "a" and the increase from point to point "r" then the values are a, a+ r, a+ 2r, ..., a+ 99r. The fact that the final amount, a+ 99r, is 3.3 times the initial amount, a, means that a+ 99r= 3.3a.

Subtracting a from both sides, 99r= 2.3a.

We also have the fact that we are "spreading out 1000" which I take to mean that these numbers add to 1000: a+ (a+ r)+ (a+ 2r)+ ...+ a+ 99r= 100a+ (1+ 2+ ...+ 99)r.= 1000.

To sum that last arithmetic series write
S= 1+ 2+ 3+ ...+ 97+ 98+ 99 and reverse the order of the numbers
S= 99+ 98+ 97+ ... + 3+ 2+ 1 . Now sum vertically.
2S= 100+ 100+ 100+ ...+ 100+ 100+ 100= 99(100)= 9900.
So S= 4950 and we have 100a+ 4950r= 1000.

Solve 99r= 2.3a and 100a+ 4950r= 1000.
r= (2.3/99)a= 0.023a so 100a+ 4950r= 100a+ 115a= 215a= 1000 and a= 1000/215= 8.69 approximately.

Then r= (2.3/99)(8.69)= 0.2020.

I will need your help and guidance with basic probability problems. I like probability but the course is a bit fuzzy.

 

[Country Boy]

I will need your help and guidance with basic probability problems. I like probability but the course is a bit fuzzy.

?? This problem doesn't have anything to do with "probability". If you want help with probabilities please pot a probability problem.