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
We have to calculate how much more money does Mike deposit each month. This can be done using the formula for the arythmetic sequence; S n = n/2 ( a1 + an ). Here: a 1 = 9, n = 41 and a n = a 1 + ( n-1 ) d = 9 + 40 d. So : 9,389 = 41/2 ( 9 + 9 + 40 d ); 9,389 = 41 / 2 ( 18 + 40 d ); 9,389 = 41 ( 9 + 20 d ) ; 9,389 = 369 + 820 d; 820 d = 9,389 - 369; 820 d = 9,020; d = 9,020 : 820; d = 11. Answer: He 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
\(S_n=\frac{n}{2} [2a_1+(n-1)d]\\\)
\(a_1=9\ =41\\d=11 \\\\S_4_1=\frac{41}{2}[ (2)(4)+(41-1)(11)] \\\\S_4_1=\frac{41}{2} [8+(40)(11)]\\\\\S_4_1=\frac{41}{2} (18+440)\\\\S_4_1=\frac{41}{2} (458)\\\\S_4_1=20.5 (458)\\\\S_4_1=9,389\)
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.