# Find the average spacing between an array of numbers

If I had an array of 7 numbers and I wanted all numbers to be equally spaced within it but needed to start at 5 and end at 35 how would I do this?

For instance if I were given:

• Numbers in array: 7
• Largest number in array: 35
• Smallest number in array: 5

How do I find what number to increment the array by if I wanted to create an array of equally spaced numbers?

An example answer would be: Given the above three numbers I would say increment each array position by 5 and all 7 numbers will be equally spaced all the way to 35. The end result would look like this: [5,10,15,20,25,30,35]

Question 1: What would the formula be to repeat this any time I have array length, smallest, and largest number?

Question 2: In the above example if I were given the number 30 what is the formula to find its position in the given array? The formula should produce a result like this: If the array started at index one it would be position 6.

Any help would be great, thanks!

• How many gaps between consecutive pairs of numbers? In your example, $$7-1=6$$
• What do the gaps add up to? In your example, $$35-5=30$$
• What is the size of a gap? In your example, $$\frac{30}{6}=5$$
• suppose there was a value before the first; what would it be? (call this the zeroth value) In your example, $$5-5=0$$
• how many gaps would you need to add to the zeroth value to get your target value? In your example, $$\frac{30}{5}=6$$
So the $$n$$th value would be the zeroth value plus $$n$$ gaps, while value $$x$$ would be in the position corresponding to $$x$$ minus the zeroth value, all divided by the gap size
We have 2 numbers, $$a,b$$ with $$a. Suppose we want $$c$$ numbers in the array. Then the array would look like this: $$(a,a+d,a+2d,....,a+(c-2)d,b)$$ For some number $$d$$. Notice that the difference between any consecutive numbers in the array is $$d$$ (by construction). Using this fact, $$b-(a+(c-2)d)=d.$$ (last number in array minus penultimate) which we can solve for $$d$$ to obtain $$d=(b-a)/(c-1)$$ and $$c \neq 1$$ since we always have at least 2 numbers in the array.