Question 982715
For a geometric sequence with first term {{{b[1]}}} and common ratio {{{r}}} ,
the sum of the first {{{n}}} terms is
{{{S[n]=b[1]+b[1]*r+b[1]*r^2+b[1]*r^3+"..."+b[1]*r^(n-2)+b[1]*r^(n-1)}}}
{{{S[n]=b[1]*(1+r+r^2+r^3+"..."+r^(n-2)+r^(n-1))=b[1]*((r^n-1)/(r-1))=b[1]*((1-r^n)/(1-r))}}}
When {{{r<1}}} , {{{r^n}}} gets progressively smaller as {{{n}}} increases, approaching zero,
so the sum of an infinite geometric series with first term {{{b[1]}}} and common ratio {{{r<1}}} is
{{{S=b[1]*(1/(1-r))}}}
In this case,
{{{S=70*(1/(1-(-0.23)))=70/(1+0.23)=70/1.23=highlight(56.91)}}} rounded to two decimal places.