Question 1154077
he second term of a GP is 4, the fifth term is 81. Find the seventh term

{{{a[2]=4}}}
{{{a[5]=81}}}


we need to find first term and common ratio 

using nth term formula of a GP :


{{{a[n] = a[1]*r^(n-1)}}} where {{{a[1]}}} is first term, {{{r}}} is common ratio, and {{{n}}} is a number of terms 


given:


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

{{{4= a[1]*r}}}

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


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

{{{81= a[1]*r^4}}}

{{{r^4=81/a[1]}}}

{{{r=root(4,81/a[1])}}}....eq.2



from eq.1 and eq.2 we have


{{{4/a[1]=root(4,81/a[1])}}}....take to fourth power


{{{4^4/a[1]^4=81/a[1]}}}


{{{256/a[1]^4=81/a[1]}}}


{{{256=81a[1]^4/a[1]}}}


{{{256=81a[1]^3}}}


{{{256/81=a[1]^3}}}


{{{a[1]^3=256/81}}}


{{{a[1]=root(3,256/81)}}}


{{{a[1]=root(3,256)/root(3,81)}}}


{{{a[1]=4root(3,4)/3root(3,3)}}}


{{{a[1]=(4/3)(root(3,4)/root(3,3))}}}


go to


{{{r=4/a[1]}}}....eq.1, substitute {{{a[1]}}}


{{{r=4/(4/3)(root(3,4)/root(3,3))}}}


{{{r=12/(4(root(3,4)/root(3,3)))}}}


{{{r=(3/2)*root(3, 6)}}}



{{{a[n] = (4/3)(root(3,4)/root(3,3))*((3/2)*root(3, 6))^(n-1)}}}


7th term: n=7


{{{a[7] = (4/3)(root(3,4)/root(3,3))*((3/2)*root(3, 6))^(7-1)}}}


{{{a[7] = (4/3)(root(3,4)/root(3,3))*((3/2)*root(3, 6))^6}}}


{{{a[7] = 1.46752322173*410.0625}}}


{{{a[7] =601.8}}}