Question 1083649
The first term is {{{b[1]=5}}} .
The common ratio is r=15÷5=3 .
Term number {{{n}}} is {{{b[n]=5*3^(n-1)}}} .
The sum of the first {{{n}}} terms is
{{{b[1](r^n-1)/(r-1)=5(3^n-1)/(3-1)=5.(3^n-1)/2}}}
If we want to find what is the smallest n
that will make the,sun of the first n terms no less than 100, our inequality is
{{{5(3^n-1)/2>=1000}}}
{{{3^n-1>=1000*2/5}}}
{{{3^n-1>=400}}}
{{{3^n>=401}}}
{{{3^6=729}}}
{{{3^5=243}}}
The sum of first {{{n}}} terms that is no less than 1000 is the sum of the first 6 terms:
5+15+45+135+405+1215.
The sum of some terms cannot be exactly 1000.
For the sum of some terms of the form {{{b[k+1]=5*3^k}}}
(with {{{k>=0}}} being some integer)
to be exactly 1000,
the sum of all the {{{3^k}}} factors has to be 200.´
All those factors are multiples of 3, except {{{3^0=1}}}.
and the sum of multiples of 3 cannot be either 200 or 199.