Question

(08.02) Mike started a savings account by depositing $9. Each month, he deposits more money than the month before. At the end of 41 months, he has saved $9,389.00. How much more does he deposit each month?

Answer 1

$11.00 or 11

same thing :)

*plug all other answer choices inside formula as d to check your answer*

Explanation:

we need to use the arithmetic sequence formula

`S_n=(n)/(2) [2a_1+(n-1)d]\\`

`a_1=9\\n=41\\d=11 \\\\S_4_1=(41)/(2)[ (2)(4)+(41-1)(11)] \\\\S_4_1=(41)/(2) [8+(40)(11)]\\\\\S_4_1=(41)/(2) (18+440)\\\\S_4_1=(41)/(2) (458)\\\\S_4_1=20.5 (458)\\\\S_4_1=9,389`

He deposit 11 each month is right because the sum of the series (
`S_n`) is 9,389 and he saved 9,389 at the end of the 41 months.

Answer 2

Mike deposits $11 more each month.

Explanation:

Since, every month Mike is paying more money then the amount of money deposited can be in the form of an Arithmetic Progression.

First term, a = 9

Total number of months, n = 41

Total sum of money, Sn = $9389

We need to find amount which Mike is paying more every month that is d or common difference of the resulting A.P.

`S_n=(n)/(2)* (2\cdot a+(n-1)* d)\\\\9389=(41)/(2)* (2\cdot 9+40\cdot d)\\\\229* 2=18+40\cdot d\\\\\implies 40\cdot d=458-18\\\\\implies 40\cdot d = 440\\\\\implies d=11`

Hence, Mike deposits $11 more each month.