Respuesta :
Answer:
a.The total amount that Molly will have in her account at the end of 3 years is $345.60
b. The sum of finite G.P series , [tex] S_n= \frac{a(1-r^n)}{1-r} [/tex]when r<1
The sum of finite G.P series ,[tex]S_n=\frac{a(r^n-1)}{r-1}[/tex] when r>1
Step-by-step explanation:
Given
Molly deposits saving in her account= $200
Interest rate=20%
Time =3 years
Amount,A[tex]=P(1+\frac{r}{100})^t[/tex]
Where P= principle value
r= Interest rate annually
t= time in years
A=Amount
P=$200
r=20%
t=3 years
Substitute all values in the given formula
A=[tex]200(1+\frac{20}{100})^3[/tex]
A=[tex] 200\times \frac{6}{5}\times\frac{6}{5}\times\frac{6}{5}[/tex]
A=[tex]\frac{8\times 6\times 6\times 6}{5}[/tex]
A=[tex]\frac{1728}{5}[/tex]
A=345.6
Amount=$345.6
Hence, Molly will have total amount in her account at the end of 3 years is $345.6.
b. Let finite G. P series
[tex]a,ar,ar^2,ar^3,..........ar^{n-1}[/tex]
Total terms=n
[tex]a_1=a,a_2=ar,a_3=ar^3[/tex]
[tex]\frac{a_2}{a_1}=\frac{ar}{a}=r[/tex]
[tex]\frac{a_3}{a_2}=\frac{ar^2}{ar}=r[/tex]
Hence, common ratio=r
Sum of finite G.P series,[tex]S_n=\frac{a(r^n-1)}{r-1}[/tex] when r>1
Sum of finite G.P series,[tex]S_n=\frac{a(1-r^n)}{1-r}[/tex] when r<1.
