Question

Kathryn deposits 100 into an account at the beggining of each 4 year period for 40 years. The account credits interest at an effective annual interest rate of i. The accumulated amount in the account at the end of 40 years is X. This is 5 times the accumulated amount in the account at the ned of 20 years. Calculate X

Answer 1

Answer:

X = 6,194.84

Step-by-step explanation:

Step 1

The value of X is:

$X = 100 [(1+i)^{40}+(1+i)^{36}+...+(1+i)^{4}]$

$X=\frac{100(1+i)^{4}[1-(1+i)^{40}]}{1-(1+i)^{4} }$

Step 2

The value of Y is:

Y = A x (20)

$Y = 100 [(1+i)^{20}+(1+i)^{16}+...+(1+i)^{4}]$

$Y=\frac{100(1+i)^{4}[1-(1+i)^{20}]}{1-(1+i)^{4} }$

Step 3

We are informed that X = 5Y. Therefore,

$1 + (1+i)^{20}=5$

$(1+i)^{20}=4$

$(1+i)^{4}=4^{1/5}$

$(1+i)^{4}=1.3195$

Step 4

Substituting the above value in X = 5Y:

$X=\frac{100(1.3195)(1-4^{2} )}{1-1.3195}$

X = 6,194.84

Answer 2

Answer:

6194.84

Step-by-step explanation:

Using the formula for calculating accumulated annuity amount

F = P × ([1 + I]^N - 1 )/I

Where P is the payment amount. I is equal to the interest (discount) rate and N number of duration

For 40 years,

X = 100[(1 + i)^40 + (1 + i)^36 + · · ·+ (1 + i)^4]

=[100 × (1+i)^4 × (1 - (1 + i)^40]/1 − (1 + i)^4

For 20 years,

Y = A(20) = 100[(1+i)^20+(1+i)^16+· · ·+(1+i)^4]

Using X = 5Y (5 times the accumulated amount in the account at the ned of 20 years) and using a difference of squares on the left side gives

1 + (1 + i)^20 = 5

so (1 + i)^20 = 4

so (1 + i)^4 = 4^0.2 = 1.319508

Hence X = [100 × (1 + i)^4 × (1 − (1 + i)^40)] / 1 − (1 + i)^4

= [100×1.3195×(1−4^2)] / 1−1.3195

X = 6194.84