We have that
[tex]\begin{gathered} Sn=a+ar^1+ar^2+...+ar^{n-1} \\ S_2=3.5,S_3=11.375 \end{gathered}[/tex]Now, we have two equations:
[tex]\begin{gathered} a+ar=3.5 \\ a+ar^1+ar^2=11.375 \end{gathered}[/tex]We have that:
[tex]S_n=\frac{a\mleft(1-r^n\mright)}{1-r}[/tex]Then, for S2 and S3:
[tex]\begin{gathered} S_2=\frac{a(1-r^2)}{1-r}=3.5 \\ S_3=\frac{a(1-r^3)}{1-r}=11.375 \end{gathered}[/tex]Then, we have two equations:
[tex]\begin{gathered} \frac{a(1-r^2)}{1-r}=3.5 \\ \frac{a(1-r^3)}{1-r}=11.375 \end{gathered}[/tex]Now, solving for (1 - r):
[tex]\begin{gathered} \frac{a(1-r^2)}{3.5}=1-r \\ \frac{a(1-r^3)}{11.375}=1-r \end{gathered}[/tex]We equate them:
[tex]\frac{a(1-r^2)}{3.5}=\frac{a(1-r^3)}{11.375}[/tex]Now, we can solve for r:
[tex]\begin{gathered} \frac{a(1-r^2)}{3.5}=\frac{a(1-r^3)}{11.375} \\ \downarrow \\ \frac{1-r^2}{3.5}=\frac{1-r^3}{11.375} \\ \downarrow \\ 11.375\cdot(1-r^2)=3.5(1-r^3) \end{gathered}[/tex]Dividing both sides by 3.5:
[tex]\begin{gathered} 11.375\cdot(1-r^2)=3.5(1-r^3) \\ \downarrow \\ 3.25(1-r^2)=1-r^3 \\ \downarrow \\ 3.25-3.25r^2=1-r^3 \end{gathered}[/tex]Now, rearraging the equation we have that:
[tex]r^3-3.25r^2+2.25=0[/tex]From the equation, we have that r has three possible values:
[tex]\begin{gathered} r_1=3 \\ r_2=_{}1 \\ r_3=-\frac{3}{4}=-0.75 \end{gathered}[/tex]We know this because of the graph of the equation:
Since r < 0, then the solution is:
[tex]r=-\frac{3}{4}=-0.75[/tex]Since
[tex]\begin{gathered} a+ar=3.5 \\ \downarrow \\ a(1+r)=3.5 \end{gathered}[/tex]we can find a if we replace r = -3/4:
[tex]\begin{gathered} a(1-0.75)=3.5 \\ a(0.25)=3.5 \end{gathered}[/tex]Then,
[tex]a=\frac{3.5}{0.25}=14[/tex]Therefore the values of r and a are:
Answer- r = - 0.75 and a = 14
