Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2005-08-04 18:01:23

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?

#2 2005-08-04 18:24:48

ganesh
Moderator
Registered: 2005-06-28
Posts: 14,803

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?


Character is who you are when no one is looking.

Offline

Board footer

Powered by FluxBB