Math Is Fun Forum

scott

Guest

three today three more tomorrow....

if I trust you for 3 min today and three min more tomorrow and thre more min the next day, how do I write a formula to figure out how long I have trusted you for in , lets say, 23 days?

Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 46,182

Re: three today three more tomorrow....

This is an Arithmetic Progression (AP).
An Arithmetic Progression is something like
1,2,3,4,5,6,7,8 or
2,4,6,8,10,12, or
9,12,15,18,21,24 or
16, 13, 10, 7...........
where the difference between two successive terms is the same.
This is called 'd' or common difference.
If 'a' is the first term of an AP and 'd' is the common difference,
the nth term would be a+(n-1)d
In your case, d is 3 minutes. n is 23, and a is also 3.
Therefore, on the 23rd day, you have trusted me for
3 + (23-1)3 = 3 + (22)3 = 3 + 66 = 69 minutes.

If you require the sum of all the terms, the formula is
n/2 [2a+(n-1)d]
In the example given by you, it would be
23/2 [6 + 22(3)] = 23/2[6+66] = 23/2[72]=23*36 = 828 minutes
Is that clear?

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