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Question
Find accumulated value after 1 year of an annuity immediate in which ₹ 10,000 is invested every quarter at 16% p.a. compounded quarterly. [Given (1.04)4 = 1.1699]
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Solution
Given, C = ₹ 10,000
Amount is invested every quarter for one year.
∴ n = 4
Rate of interest is 16% p.a
∴ r = `(16)/(4)` = 4%
i = `"r"/(100) = (4)/(100)` = 0.04
Since, A = `"C"/"i"[(1 + "i")^"n" - 1]`
= `(10,000)/(0.04)[(1 + 0.04)^4 - 1]`
= `(10,000 xx 100)/(0.04 xx100)[(1.04)^4 - 1]`
= `(10,00,000)/4(1.1699 - 1)`
= 2,50,000 × 0.1699
= 42,475
∴ Amount accumulated value after 1 years is ₹ 42,475.
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[Given (1.1)4 = 1.4641]
For annuity due,
C = ₹ 20,000, n = 3, I = 0.1, (1.1)–3 = 0.7513
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`
For an annuity due, C = ₹ 2000, rate = 16% p.a. compounded quarterly for 1 year
∴ 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
