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Question
Find the accumulated value of annuity due of ₹1,000 p.a. for 3 years at 10% p.a. compounded annually. [Given (1.1)3 = 1.331]
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Solution
Given C = ₹1,000, n = 3 years, r = 10% p.a.
∴ i = `"r"/(100) = (10)/(100)` = 0.1
Now, A = `("C"(1 + "i"))/"i"[(1 + "i")^"n" - 1]`
∴ A = `(1,000(1 + 0.1))/(0.1)[(1 + 0.1)^3 - 1]`
= `(1,000(1.1))/(0.1)[(1.1)^3 - 1]`
= (1,000)(11)[1.331 – 1]
= 11,000(0.331)
∴ A = 3,641
∴ Accumulated value of annuity due is ₹3,641.
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For annuity due,
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Therefore, P = `square/0.1 xx [1 - (1 + 0.1)^square]`
= 2,00,000 [1 – 0.7513]
= ₹ `square`
The future amount, A = ₹ 10,00,000
Period, n = 20, r = 5%, (1.025)20 = 1.675
A = `"C"/"I" [(1 + "i")^"n" - 1]`
I = `5/200` = `square` as interest is calculated semi-annually
A = 10,00,000 = `"C"/"I" [(1 + "i")^"n" - 1]`
10,00,000 = `"C"/0.025 [(1 + 0.025)^square - 1]`
= `"C"/0.025 [1.675 - 1]`
10,00,000 = `("C" xx 0.675)/0.025`
C = ₹ `square`
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∴ Rate of interest per quarter = `square/4` = 4
⇒ r = 4%
⇒ i = `square/100 = 4/100` = 0.04
n = Number of quarters
= 4 × 1
= `square`
⇒ P' = `(C(1 + i))/i [1 - (1 + i)^-n]`
⇒ P' = `(square(1 + square))/0.04 [1 - (square + 0.04)^-square]`
= `(2000(square))/square [1 - (square)^-4]`
= 50,000`(square)`[1 – 0.8548]
= ₹ 7,550.40
