Question 1203500
given:
 
{{{-1}}},{{{2}}},{{{-4}}},{{{8}}},{{{-16}}}

 
since differences between two consecutive  terms are not same, sequence is not arithmetic

check if it is geometric sequence

A sequence an where there is a fixed r so that{{{ a^n/a^(n-1)=r}}} for all {{{n}}} is called a geometric sequence. The number {{{r}}} is usually called the ratio. 

use given terms 

{{{-1 }}}and {{{2}}}

{{{2/-1=r }}}
{{{-2=r }}}

{{{2 }}}and {{{-4}}}

{{{-4/2=r }}}
{{{-2=r }}}


 The formula for the {{{n}}}th term of this sequence is: 


{{{a[n]= a[1]*r^(n-1)}}}


since {{{a[1]=-1 }}}and {{{r=-2}}}, we have


{{{highlight(a[n]= -1*(-2)^(n-1))}}}



check the formula:

given that fourth term is  {{{a[4]=8}}}, where {{{n=4}}}

{{{8= -1*(-2)^(4-1)}}}

{{{8= -1*(-2)^3}}}

{{{8= -1*(-8)}}}

{{{8=8 }}} which confirms our formula