Question

Bob makes his first $ 1, 100 deposit into an IRA earning 6.8 % compounded annually on his 24th birthday and his last $1, 100 deposit on his 36th birthday ​(13 equal deposits in​ all). With no additional​ deposits, the money in the IRA continues to earn 6.8 % interest compounded annually until Bob retires on his 65th birthday. How much is in the IRA when Bob​ retires?

Answer 1

Answer:

$133991.2

Step-by-step explanation:

Bob makes his first $1000 deposit into an IRA earning 6.8% compounded annually on his 24th birthday and his last $1000 deposit on his 36th birthday (13 equal deposits in all).

Therefore, till his retirement on his 65th birthday, the first deposit of $1000 will compound for (65 - 24) = 41 years.

His second deposit of $1000 will compound for 40 years and so on up to his 13th deposit of $1000, which will be compounded for ( 65 - 36) = 29 years.

Therefore, after retirement in his IRA there will be total

$$[1000(1 + \frac{6.8}{100} )^{41} + 1000(1 + \frac{6.8}{100} )^{40} + 1000(1 + \frac{6.8}{100} )^{39} + ........ + 1000(1 + \frac{6.8}{100} )^{29}]$ dollars

= $$1000[1.068^{41} + 1.068^{40} + 1.068^{39} + .......... + 1.068^{29}]$

So, this is a G.P. whose number of terms is 13, the first term is $1.068^{29}$ and common ratio is 1.068, then using formula for sum of G.P. we get,

= $$1000\times (1.068)^{29} [\frac{(1.068)^{13} - 1}{1.068 - 1} ]$

= $$(1000 \times 6.74 \times 19.88)$

= $133991.2 (Answer)