Question 1127066
 the third term of a geometric sequence is {{{4}}}
 {{{a[3]=4}}}
 and the sixth term is {{{32/27}}}
{{{a[6]=32/27}}}

find the nth term:

The formula for the general term for each geometric sequence is {{{a[n]=a[1]*r^(n-1)}}} where {{{a[1]}}} is first term and {{{r}}} is common ratio

we need to find {{{a[1]}}} and {{{r}}}


since given {{{a[3]=4}}}, we know that {{{n=3}}} then


{{{a[3]=a[1]*r^(n-1)}}}

{{{4=a[1]*r^(3-1)}}}

{{{4=a[1]*r^2}}}....solve for {{{a[1]}}}

{{{a[1] =4/r^2}}}........eq.1


since given {{{a[6]=32/27}}}, we know that {{{n=6}}} then

{{{a[6]=a[1]*r^(6-1)}}}

{{{32/27=a[1]*r^5}}}.....solve for {{{a[1]}}}

{{{a[1] =(32/27)/r^5}}}

{{{a[1] =32/(27r^5)}}}.......eq.2


from eq.1 and eq.2 we have


{{{32/(27r^5)=4/r^2}}}....solve for {{{r}}}

{{{32r^2=4*27r^5}}}

{{{cross(32)8*r^2=cross(4)27r^5}}}

{{{8r^2=27r^5}}}

{{{r^5/r^2=8/27}}}

{{{r^3=8/27}}}

{{{r^3=2^3/3^3}}}

{{{r^3=(2/3)^3}}}

{{{r=(2/3)}}}


now find {{{a[1]}}}

{{{a[1] =4/r^2}}}........eq.1

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

{{{a[1] =4/(4/9)}}}

{{{a[1] =(9*4)/4}}}

{{{a[1] =9}}}


the nth term formula is: {{{a[n]=9*(2/3)^(n-1)}}}