Question

5. Calculate the compound interest for the second year on 6000 invested for 3 years
10% p.a. Also find the sum due at the end of third year.​

Answer 1

Step-by-step explanation:

A=P(1+r/n)^(nt)

A=final amount

P=initial amount

n=number of time interest applied per time period

r=interest rate

t=number of time period elapsed

after 1 year

A = 6000(1+10/100)

= 6000(1+1/10)

= 6000(1+0.1)

= 6000(1.1)

A = 6600

after 2 year

A = (1.1)6000(1+10/100)

= (1.1)6000(1+1/10)

= (1.1)6000(1+0.1)

= (1.1)6000(1.1)

= 6000(1.1)^2

=6000(1.21)

A =7260

after 3 year

A = (1.1)^2(6000)(1+10/100)

= (1.1)^2(6000)(1+1/10)

= (1.1)^2(6000)(1+0.1)

= (1.1)^2(6000)(1.1)

= 6000(1.1)^3

=6000(1.331)

A =7986

the sum due at the end of third year.