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
Answer:
X = 6,194.84
Step-by-step explanation:
Step 1
The value of X is:
[tex]X = 100 [(1+i)^{40}+(1+i)^{36}+...+(1+i)^{4}][/tex]
[tex]X=\frac{100(1+i)^{4}[1-(1+i)^{40}]}{1-(1+i)^{4} }[/tex]
Step 2
The value of Y is:
Y = A x (20)
[tex]Y = 100 [(1+i)^{20}+(1+i)^{16}+...+(1+i)^{4}][/tex]
[tex]Y=\frac{100(1+i)^{4}[1-(1+i)^{20}]}{1-(1+i)^{4} }[/tex]
Step 3
We are informed that X = 5Y. Therefore,
[tex]1 + (1+i)^{20}=5[/tex]
[tex](1+i)^{20}=4[/tex]
[tex](1+i)^{4}=4^{1/5}[/tex]
[tex](1+i)^{4}=1.3195[/tex]
Step 4
Substituting the above value in X = 5Y:
[tex]X=\frac{100(1.3195)(1-4^{2} )}{1-1.3195}[/tex]
X = 6,194.84
