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Question
Find the amount of an ordinary annuity if a payment of ₹ 500 is made at the end of every quarter for 5 years at the rate of 12% per annum compounded quarterly. [Given (1.03)20 = 1.8061]
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Solution
Given, C = ₹ 500
Amount is invested at the end of every quarter.
∴ It is an immediate annuity.
Rate of interest is 12% p.a.
∴ r = `(12)/(4)`% = 3% per quarter
∴ i = `"r"/(100) = (3)/(100)` 0..03
The period is of 5 years and payment is made on quarterly basis.
∴ n = 5 x 4 = 20
Since, A = `"C"/"i"[(1 + "i")^"n" - 1]`
= `(500)/(0.03)[(1 + 0.03)^20 - 1]`
= `(500)/(0.03)[(1.03)^20 - 1]`
= `(500)/(0.03)(1.8061 - 1)`
= `(500)/(0.03) xx (0.8061)`
= `(403.05)/(0.03)`
= `(40305)/(3)`
= ₹ 13,435
∴ Amount of ordinary annuity is ₹ 13,435.
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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
