Question 1190303


general formula for nth term:

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


if  {{{10}}}th term of a G.P is {{{2^10}}}, we have


{{{2^10=a*r^(10-1)}}} 

{{{2^10=a*r^9}}}

{{{a=2^10/r^9}}}............eq.1



if the {{{20}}}th term is given as {{{2^19}}}, we have


{{{2^19=a*r^(20-1)}}}

{{{a=2^19/r^19}}}.........eq.2


from eq.1 and eq.2 we have


{{{2^10/r^9=2^19/r^19}}}.....solve for{{{ r}}}

{{{r^19/r^9=2^19/2^10}}}

{{{r^10=2^9}}}

{{{r = 2^(9/10)}}}


then

{{{a=2^10/r^9}}}............eq.1, substitute{{{ r}}}

{{{a=2^10/(2^(9/10))^9}}}

{{{a=2^10/2^(81/10)}}}

{{{a=2* 2^(9/10)}}}


so, 

first term is {{{a = 2*2^(9/10)}}} and  common ratio is {{{r = 2^(9/10)}}}


your nth term formula is:


{{{a[n]= 2*2^(9/10)*( 2^(9/10))^(n-1)}}}