Question

At the beginning of every year, Molly deposits $200 in a savings account that offers an interest rate of 20%, compounded annually. The total amount that Molly will have in her account at the end of 3 years is $

Find the sum of a finite geometric series.

Answer 1

She would have $873.60 dollars at the end of 3 years

Answer 2

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 ,
`S_n= (a(1-r^n))/(1-r)`when r<1

The sum of finite G.P series ,
`S_n=(a(r^n-1))/(r-1)` when r>1

Explanation:

Given

Molly deposits saving in her account= $200

Interest rate=20%

Time =3 years

Amount,A
`=P(1+(r)/(100))^t`

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=
`200(1+(20)/(100))^3`

A=
`200* (6)/(5)*(6)/(5)*(6)/(5)`

A=
`(8* 6* 6* 6)/(5)`

A=
`(1728)/(5)`

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

`a,ar,ar^2,ar^3,..........ar^(n-1)`

Total terms=n

`a_1=a,a_2=ar,a_3=ar^3`

`(a_2)/(a_1)=(ar)/(a)=r`

`(a_3)/(a_2)=(ar^2)/(ar)=r`

Hence, common ratio=r

Sum of finite G.P series,
`S_n=(a(r^n-1))/(r-1)` when r>1

Sum of finite G.P series,
`S_n=(a(1-r^n))/(1-r)` when r<1.