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Question
A person plans to put ₹400 at the beginning of each year for 2 years in a deposit that gives interest at 2% p.a. compounded annually. Find the amount that will be accumulated at the end of 2 years.
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Solution
Given, C = ₹400, n = 2 years, r = 2% p.a.
i = `"r"/(100) = (2)/(100)` = 0.02
Now,A = `("C"(1 + "i"))/"i"[(1 + "i")^"n" - 1]`
∴ A = `(400(1 + 0.02))/(0.02)[(1 + 0.02)^2 - 1]`
= `(400(1.02))/(0.02)[(1.02)^2 - 1]`
= (400)(51)[1.0404 – 1]
= 20,400(0,0404)
A = 824.16
∴ Accumulated amount after 2 years is ₹824.16.
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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`
