# Find an increasing arithmetic sequence of n real numbers whose sum is equal to N

For example, i am looking for the sequence of 7 increasing numbers starting from 0 whose sum is 0.7 (more or less) with uniform distribution (the gap between the numbers are equal). It seems to me there is something to do with the pyramid but I did not find what i wanted into the web.

edit: this kind of sequence is usually called arithmetic progression

• This is not clear. seven numbers sum to a fixed number, where does a distribution enter into it? – lulu Aug 7 '18 at 13:41
• starting from 0 – elldekaa Aug 7 '18 at 13:43
• That doesn't help. Can you give an example of a sequence of $7$ numbers that sums to $.7$ but does not meet your criterion, and explain how it fails? – saulspatz Aug 7 '18 at 13:44
• It is still unclear what you mean. Would 7x0.1 work? If not, why not? – Ingix Aug 7 '18 at 13:45
• As I said, this isn't clear at all. Did you want the gaps to be equal? You didn't specify that. If we call the gap $g$ then your sequence is $0,g,2g,3g,4g,5g,6g$ which sums to $g\times (1+2+3+4+5+6)=21g$. If you want that to be $.7$ we solve as $g=\frac {.7}{21}=.0\overline 3$. Is that what you wanted? – lulu Aug 7 '18 at 13:53

Given $n\in \mathbb N$ the OP is asking for a value $g_n$ such that the series $$0+g_n+2g_n+\cdots +(n-1)g_n=.7$$
We remark that this can be solved as $$.7=g_n\times (1+2+\cdots +(n-1))=g_n\times \frac {n(n-1)}2\implies \boxed {g_n=\frac {1.4}{n(n-1)}}$$
For instance, with $n=7$ we get $g_7=\frac 1{30}$.
Of course, there is nothing special about the value $.7$