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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.
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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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
