Respuesta :
Answer:
Mike deposits $11 more each month.
Step-by-step 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.
[tex]S_n=\frac{n}{2}\times (2\cdot a+(n-1)\times d)\\\\9389=\frac{41}{2}\times (2\cdot 9+40\cdot d)\\\\229\times 2=18+40\cdot d\\\\\implies 40\cdot d=458-18\\\\\implies 40\cdot d = 440\\\\\implies d=11[/tex]
Hence, Mike deposits $11 more each month.
Answer:
$11.00 or 11
same thing :)
*plug all other answer choices inside formula as d to check your answer*
Step-by-step explanation:
we need to use the arithmetic sequence formula
[tex]S_n=\frac{n}{2} [2a_1+(n-1)d]\\[/tex]
[tex]a_1=9\\n=41\\d=11 \\\\S_4_1=\frac{41}{2}[ (2)(4)+(41-1)(11)] \\\\S_4_1=\frac{41}{2} [8+(40)(11)]\\\\\S_4_1=\frac{41}{2} (18+440)\\\\S_4_1=\frac{41}{2} (458)\\\\S_4_1=20.5 (458)\\\\S_4_1=9,389[/tex]
He deposit 11 each month is right because the sum of the series ([tex]S_n[/tex]) is 9,389 and he saved 9,389 at the end of the 41 months.
