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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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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`
