# Application of Geometric Series Problem (1 Viewer)

#### duckylah21

##### New Member
http://ibb.co/myefJ

I’m having trouble with Q3 and Q4.

Q3) I’ve done multiple tries like dividing her yearly $100 by 12 months and still can’t get the right answer. It is compounded monthly but she only invests once at the beginning of each year. Q4) Got close to the answer. Answers: Q3)$5840.77
Q4) $9732.09 ($1177.26 + $8554.83) Any help is greatly appreciated. Thank you! #### integral95 ##### Well-Known Member http://ibb.co/myefJ I’m having trouble with Q3 and Q4. Q3) I’ve done multiple tries like dividing her yearly$100 by 12 months and still can’t get the right answer. It is compounded monthly but she only invests once at the beginning of each year.

Q4) Got close to the answer.

Q3) $5840.77 Q4)$9732.09 ($1177.26 +$8554.83)

Any help is greatly appreciated. Thank you!
Yeah dividing that amount doesn't do anything

So for Q3 it goes like this, basically after every 12 months, \$100 gets added into the account

$\bg_white A_{12} = 100(1.0075)^{12} \\ \\ A_{13} = (100+A_{12})(1.0075) \\ \\ \dots \\ A_{24} = (100+A_{12})(1.0075) ^{12} = 100(1.0075)^{12} + 100(1.0075)^{24}$

You'll then notice this pattern continues until you reach 240 months (20 years), thus giving

$\bg_white A_{240} = 100((1.0075)^{12} + (1.0075)^{24} + \dots (1.0075)^{240}) = \\ \\ 100(1.0075)^{12}(\frac{(1.0075)^{240}-1}{(1.0075)^{12}-1}) = \\ \\ 5840.76939$