Question 1044698

To find the sum of a certain number of terms of a geometric sequence:

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

where {{{S[n]}}}is the sum of {{{n}}} terms (nth partial sum), 
 {{{a[1]}}} is the first term,  {{{r}}} is the common ratio.

if given the first term is  {{{a[1]=4}}}, the last term is {{{a[n]=324}}}, and the common ratio is {{{r=3}}} than we need first to find what  number is the last term:

To find any term of a geometric sequence:
{{{a[n]=a[1]*r^(n-1)}}}
where a[1] is the first term of the sequence, {{{r }}}is the common ratio, {{{n}}} is the number of the term to find

{{{324=4*3^(n-1)}}}

{{{324/4=3^(n-1)}}}

{{{81=3^(n-1)}}}

{{{3^4=3^(n-1)}}}.........since base same, exponents must be same too, so we have

{{{4=n-1}}}

{{{n=5}}}-> so, your sequence have {{{5}}} terms

than the sum will be:

{{{S[5]=4(1-3^5)/(1-3)}}}

{{{S[5]=4(1-243)/(-2)}}}

{{{S[5]=2(-242)/(-1)}}}

{{{S[5]=2(242)}}}

{{{S[5]=484}}}