Question 1135685



{{{108}}}, {{{-18}}}, {{{3}}}, {{{-1/2}}}

The formula for nth term of an geometric progression is {{{a[n]=a[1]*r^(n-1)}}} 

divide two consecutive terms to find ratio {{{r}}}

{{{r=-18/108=-1/6}}}
{{{r=3/-18=-1/6}}}

so, in this example we have {{{a[1]=108}}} and {{{r=-1/6}}} so:

then


{{{a[n]=108*(-1/6)^(n-1)}}}

or

{{{a[n]=(-1)^(n + 1) *2^(3 - n)* 3^(4 - n)}}}


{{{lim(n->infinity)}}}{{{ (-1)^(-1 + n) 2^(3 - n) 3^(4 - n) = 0}}}


so, answer is: Converges; {{{0}}}