Answer to Question #216125 in Management for Vicky

Question

Question :How much amount is required to be invested every year so as to accumulate 300000 at the end of 10 year if interest is compounded annually at 10% ?

Answer 1

Solution:

A = P(1 + R/100)ⁿ

Amount Invested every Year = P

Amount invested at beginning of 1st Year will become = P(1 + R/100)¹⁰ after 10 years

Amount invested at beginning of 2nd Year will become = P(1 + R/100)⁹ after 10 years

Total amount = P(1 + R/100)¹⁰ + P(1 + R/100)⁹ + . + P(1 + R/100)² + P(1 + R/100)

This is an GP

a = P(1 + R/100)

r = (1 + R/100)

n = 10

Sₙ = a(rⁿ - 1)/( r - 1)

= P(1 + R/100) ((1 + R/100)¹⁰ - 1) / ( 1 + R/100 - 1 )

= P((100 + R)/100) ((1 + R/100)¹⁰ - 1) / ( R/100)

= P((100 + R)) ((1 + R/100)¹⁰ - 1) /R

R = 10

300000 = P((100 + 10)) ((1 + 10/100)¹⁰ - 1) /10

=> 300000 =11P ( 1.1¹⁰ - 1)

=> P = 300000/ 11 ( 1.1¹⁰ - 1)

=> P = 17,112.38