Annual pension

Calculate the amount of money generating an annual pension of EUR 1000, payable at the end of the year and for a period of 10 years, shall be inserted into the bank to account with an annual interest rate of 2%

Result

x =  8998.99 Eur

Solution:

q=1.02  x1 x2=qx1−1000 x3=qx2−1000 .... x10=qx9−1000  x10=1000  x9=(x10+1000)/q=(1000+1000)/1.02≐1960.7843 Eur x8=(x9+1000)/q=(1960.7843+1000)/1.02≐2902.7297 Eur x7=(x8+1000)/q=(2902.7297+1000)/1.02≐3826.2056 Eur x6=(x7+1000)/q=(3826.2056+1000)/1.02≐4731.5741 Eur x5=(x6+1000)/q=(4731.5741+1000)/1.02≐5619.1903 Eur x4=(x5+1000)/q=(5619.1903+1000)/1.02≐6489.4023 Eur x3=(x4+1000)/q=(6489.4023+1000)/1.02≐7342.5512 Eur x2=(x3+1000)/q=(7342.5512+1000)/1.02≐8178.9718 Eur x1=(x2+1000)/q=(8178.9718+1000)/1.02≐8998.992 Eur  x=x1=8998.992≐8998.992=8998.99  Eur q = 1.02 \ \\ \ \\ x_{ 1 } \ \\ x_{ 2 } = q x_{ 1 } - 1000 \ \\ x_{ 3 } = q x_{ 2 } - 1000 \ \\ .... \ \\ x_{ 10 } = q x_{ 9 } - 1000 \ \\ \ \\ x_{ 10 } = 1000 \ \\ \ \\ x_{ 9 } = (x_{ 10 } + 1000) / q = (1000 + 1000) / 1.02 \doteq 1960.7843 \ Eur \ \\ x_{ 8 } = (x_{ 9 } + 1000) / q = (1960.7843 + 1000) / 1.02 \doteq 2902.7297 \ Eur \ \\ x_{ 7 } = (x_{ 8 } + 1000) / q = (2902.7297 + 1000) / 1.02 \doteq 3826.2056 \ Eur \ \\ x_{ 6 } = (x_{ 7 } + 1000) / q = (3826.2056 + 1000) / 1.02 \doteq 4731.5741 \ Eur \ \\ x_{ 5 } = (x_{ 6 } + 1000) / q = (4731.5741 + 1000) / 1.02 \doteq 5619.1903 \ Eur \ \\ x_{ 4 } = (x_{ 5 } + 1000) / q = (5619.1903 + 1000) / 1.02 \doteq 6489.4023 \ Eur \ \\ x_{ 3 } = (x_{ 4 } + 1000) / q = (6489.4023 + 1000) / 1.02 \doteq 7342.5512 \ Eur \ \\ x_{ 2 } = (x_{ 3 } + 1000) / q = (7342.5512 + 1000) / 1.02 \doteq 8178.9718 \ Eur \ \\ x_{ 1 } = (x_{ 2 } + 1000) / q = (8178.9718 + 1000) / 1.02 \doteq 8998.992 \ Eur \ \\ \ \\ x = x_{ 1 } = 8998.992 \doteq 8998.992 = 8998.99 \ \text{ Eur }

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