Question 924664

{{{a[1] = -64}}},fifth term is {{{a[5]=-4}}}


a[n]=a[1]r^(n-1)...find r

{{{a[5]=a[1]r^(5-1)}}}

{{{a[5]=a[1]r^4}}}

{{{a[5]/a[1]=r^4}}}

{{{-4/-64=r^4}}}

{{{1/16=r^4}}}

{{{r=root(4,1/16)}}}

{{{r=root(4,1^4/2^4)}}}

{{{r=1/2}}}

so, formula is: {{{a[n]=a[1](1/2)^(n-1)}}}

now we can find {{{a[2]}}},{{{a[3]}}}, and {{{a[4]}}}

{{{a[2]=-64(1/2)^(2-1)}}}

{{{a[2]=-64(1/2)^1}}}

{{{a[2]=-64/2}}}

{{{a[2]=-32}}}


{{{a[3]=-64(1/2)^(3-1)}}}

{{{a[3]=-64(1/2)^2}}}

{{{a[3]=-64(1/4)}}}

{{{a[3]=-64/4}}}

{{{a[3]=-16}}}

{{{a[4]=-64(1/2)^(4-1)}}}

{{{a[4]=-64(1/2)^3}}}

{{{a[4]=-64(1/8)}}}

{{{a[4]=-64/8}}}

{{{a[4]=-8}}}


now we can write the first five terms of this geometric sequence:


{{{a[1] = -64}}}
{{{a[2]=-32}}}
{{{a[3]=-16}}}
{{{a[4]=-8}}}
{{{a[5]=-4}}}