Question 172798
Note: {{{S[5]}}} is the 5th partial sum of the geometric sequence. To find the 5th partial sum, we use the value {{{n=4}}} (since we start at {{{n=0}}}) and use the formula



{{{sum(a*r^k,k=1,n)=(a(1-r^(n+1)))/(1-r)}}}



Since we know that {{{S[5]=92.3125}}}, this means that {{{sum(a*(1.5)^n,n=1,5)=92.3125}}}



{{{92.3125=(a(1-r^(n+1)))/(1-r)}}} Replace the left side with 92.3125 



{{{92.3125=(a(1-(1.5)^(4+1)))/(1-1.5)}}} Plug in {{{r=1.5}}} and {{{n=4}}}



{{{92.3125=(a(1-(1.5)^(5)))/(-0.5)}}} Combine like terms.



{{{92.3125=(a(1-7.59375))/(-0.5)}}} Raise 1.5 to the 5th power to get 7.59375



{{{92.3125=(a(-6.59375))/(-0.5)}}} Combine like terms.



{{{92.3125=13.1875a}}} Divide



{{{92.3125/13.1875=a}}} Divide both sides by 13.1875 to isolate "a"



{{{7=a}}} Divide



So we find that {{{a=7}}} (which is the first term)



Now to find {{{S[10]}}}, we will use {{{n=9}}}



{{{(a(1-r^(n+1)))/(1-r)}}} Start with the right side of the formula



{{{(7(1-(1.5)^(9+1)))/(1-1.5)}}} Plug in {{{a=7}}}, {{{r=1.5}}}, and {{{n=9}}}



{{{(7(1-(1.5)^(10)))/(-0.5)}}} Combine like terms.



{{{(7(1-57.665))/(-0.5)}}} Raise 1.5 to the 10th power to get 57.665 (this is approximate)



{{{(7(-56.665))/(-0.5)}}} Combine like terms.



{{{(-396.655)/(-0.5)}}} Multiply



{{{793.31}}} Divide



So the tenth partial sum is approximately {{{793.31}}}. This means that {{{S[10]=793.31}}}