Question 346655
you are looking for summing an arithmetic series
--
the sequence is 10,12,14,.....    such that is sum =400
--
{{{a[i]=a[0]+(i-1)*delta}}} where {{{delta}}} is the difference between to consecutive sequence numbers,  in this case {{{a[0]=10}}} and {{{delta=2}}}
--
{{{a[i]=10+(i-1)*2}}}={{{8+2i}}}
{{{sum(a[i])}}}from 1 to n ={{{sum((8+2i))}}}= {{{sum(8)+2*sum(i)}}}={{{8n + 2*(n*(n+1)/2)}}}={{{8n+n*(n+1)}}}={{{8n+n^2+n=n^2+9n}}}
--
since we want to know when this sum=400

{{{400=n^2+9n}}}--> {{{0=n^2+9n-400}}}
--
solve for n using quadratic formula
{{{ n=(-9+-sqrt(81-4*(-400)))/2}}} 
n=-25 or n=16, since we want the number of months, we take the positive answer
--
{{{highlight(n=16 months)}}}