Question

Eric deposits 1000 into a bank account. The bank credits interest at a nominal annual rate of i convertible semiannually for the first 7 years and a nominal annual rate of 2i convertible quarterly for all years thereafter. The accumulated amount in the account at the end of 5 years is X. The accumulated amount in the account at the end of 10.5 years is 1980. Calculate X

Answer 1

1276.5

Step-by-step explanation:

Today, Eric deposited an amount(PO) that is = 1000

The annual nominal Interest rate for the first 7 years = i

This is convertible semiannually.

As such,

The semi-annual interest rate for first 7 years =
`(i)/(2)`

The number of semi-annual periods in 7 years (k) = 2*7 = 14

The magnitude of the deposit at end of year 7( P7),

= P7 == P0 * (1 + i/2)14 ------ equation 1

On the other hand,

The nominal interest rate after 7 years = 2i which is convertible quarterly

The effective quarterly interest rate =
`(2i)/(4)`=
`(i)/(2)`

The number of quarters between the end of year 7 and year 10.5 = 4* (10.5 - 7) = 14

The amount of deposit at the end of year 10.5 year = P10.5

`P_(10.5)`=
`P_(7)` * (1 +
`(i)/(2)`
`_(14)` ---equation 2

We are given the following:

`P_(10.5)` =1980

We shall thus substitute this value into the equation as

1980 =
`P_(7)` *( 1 + i/2)14

Further, we have worked out that

P7 == P0 * (1- i/2)14 *(1

We shall therefore substitute equation 1 into equation 2 as follows

1980 = P0 * (1+ i/2)14 *( 1 + i/2)14

1980 = 1000* (1+ i/2)14 *( 1 + i/2)14

This can also be rewritten as

1000* (1+ i/2)14 *( 1 + i/2)14 =1980

(1+ i/2)14 *( 1 + i/2)14 =1980/1000

(1+ i/2)28 =1.98

(1+ i/2) = 1.981/28

(1+ i/2) = 1.0247

i/2 = 1.0247 -1

i/2 = 0.0247

1 = 0.0247 *2 =0.0494

At the end of year 5, the amount has accumulated to an amount = X

X = Po * ( 1 + i/2 )10

= 1000 * ( 1 + 0.0494/2)10

= 1000 * ( 1 + 0.0247)10

= 1000 + ( 1.0247)10

= 1000* 1.276467

= 1276.5