Homework SourseA 13yearold child received an inheritance of 4000 per year T
A 13-year-old child received an inheritance of $4000 per year. This was to be invested and allowed to accumulate until the child reached 21 years of age. The first payment was made on the child\'s 13th birthday and the last on the 21st birthday. If the money was invested at 5.1% compounded annually, what did the child receive at age 21? (Round your final answer to two decimal places.)
Solution
Note we have not considered the interest received in last year. If the same is also taken into account the amount received shall be
| Dear Student | |||||||
| Thank you for using Chegg !! | |||||||
| Given | |||||||
| Amount Invested each year | = | 4000 | $ | ||||
| Rate of Interest compounded annually | = | 5.10% | |||||
| Time period | = | 8 | years | ||||
| After first year, 13th bday year, Amount is | = | 4000*(1+0.051) | (Let it be 4000x) | ||||
| =>x | = | 1.051 | |||||
| After second year, 14th bday year, Amount is | = | 4000* (X + X^2) | |||||
| After third year, 15th bday year, Amount is | = | 4000*(X + X^2 + X^3) | |||||
| Similarly after 8th year the amount will be | = | 4000*(X + X^2+X^3+ ….X^8) | |||||
| Clearly this is a geometric progression | |||||||
| with first Term (a)= 1.051 & r(Common ratio = 1.051) | |||||||
| Amount available after 8th year | = | (a)(r^n-1)/(r-1) | |||||
| = | 4000*((1.051)*((1.051^8)-1))(1.051-1) | ||||||
| = | 40288.32026 $ | ||||||
| Now child receives the above amount from his investent and 4000$ as his last inherited amount making a total of 44288.32 | |||||||
| = | 44288.32026 $ | Solution | |||||
