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

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**scott****Guest**

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?

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 18,637

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?

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

Pages: **1**