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Question
Solve the following :
Find the future value after 2 years if an amount of ₹12,000 is invested at the end of every half year at 12% p. a. compounded half yearly. [(1.06)4 = 1.2625]
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Solution
Given, C = ₹12,000
Since, the amount is invested at the end of every half year, it is immediate annuity. The period is of two years.
∴ n = 2 x 2 = 4 half years
Rate of interest is 12% p.a.
∴ r = `(12)/(2)` = 6% per half year
i = `"r"/(100) = (6)/(100)` = 0.06
Now, A = `"C"/"i"[(1 + "i")^"n" - 1]`
∴ A = `(12,000)/(0.06)[(1 + 0.06)^4 - 1]`
= 2,00,000 [(1.06)4 – 1]
= 2,00,000 (1.2625 – 1]
= 2,00,000 (0.2625)
∴ A = 52,500
∴ Future value after 2 years is ₹52,500.
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Therefore, P = `square/0.1 xx [1 - (1 + 0.1)^square]`
= 2,00,000 [1 – 0.7513]
= ₹ `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
