Question 659640
Find the nth term of the geometric sequence
125,25,5,1...{{{1/5}}},{{{1/25}}}...every next one is divided by {{{5}}}, so ratio is {{{1/5}}}

nth of General Term of a Geometric Sequence: {{{a[n]=a[1]*r^(n-1) }}}          
       
where {{{a[1]}}} is the first term of the sequence and {{{r}}} is the common ratio.

We have the common ratio {{{r=1/5}}} and {{{a[1]=125}}},then

{{{highlight(a[n]=125*(1/5)^(n-1)) }}}


so, check sixth term above using this formula:

{{{a[6]=125*(1/5)^(6-1) }}} 

{{{a[6]=125*(1/5)^5 }}} 


{{{a[6]=125*(1/25)=0.04 }}}  which is {{{1/25}}}



Find the nth term of the geometric sequence
7,49,343,2401,...
it is a "geometric sequence" which that means that each successive term is multiplied by the same factor, and in this sequence the common factor is {{{f=7}}}

 Geometric sequences have an nth term of the form {{{a[n]= a[1]f^(n-1) }}} where {{{a[1]}}} is the first term, {{{f}}} is the common factor and {{{n}}} is the number of the term (nth term).

{{{highlight(a[n]= a[1]7^(n-1)) }}} .......{{{a[1]=7}}}  and {{{f=7}}}