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Sagot :
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
[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.
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