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
$[tex][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}][/tex] dollars
= $[tex]1000[1.068^{41} + 1.068^{40} + 1.068^{39} + .......... + 1.068^{29}][/tex]
So, this is a G.P. whose number of terms is 13, the first term is [tex]1.068^{29}[/tex] and common ratio is 1.068, then using formula for sum of G.P. we get,
= $[tex]1000\times (1.068)^{29} [\frac{(1.068)^{13} - 1}{1.068 - 1} ][/tex]
= $[tex](1000 \times 6.74 \times 19.88)[/tex]
= $133991.2 (Answer)
