Question

Takuya's parents gave him \$100$100dollar sign, 100 as a birthday gift. Since he loves board games, he spent \$20$20dollar sign, 20 out of his present on board games at the end of each month until his money ran out. Let f(n)f(n)f, left parenthesis, n, right parenthesis be the amount of money that remained from Takuya's present during month nnn, where n=1n=1n, equals, 1 represents the month he received his present. fff is a sequence. What kind of sequence is it?

Answer 1

Answer:

The amount of money that remained from Takuya's present during month n is expressed as $Tn = 120-20n$ and the sequence is an ARITHMETIC SEQUENCE.

Step-by-step explanation:

If Takuya's parents gave him $100 as a birthday gift and she spent $20 each month on board games until his money ran out, this means he keeps spending $20 every month and his money keep reducing by the same amount each month until his money ran out, the following can be inferred;

initial amount = $100

If he spends $20 each month,

balance at the end of 1st month = $100-$20 = $80

balance at the end of 2nd month = $80-$20 = $60

balance at the end of  3rd month = $60-$20 = $40 and so on

The sequence formed by his balances is $100, $80, $60, $40...

Since the amount keep reducing by the same value i.e $20, then the sequence formed is an ARITHMETIC SEQUENCE.

The nth term of an arithmetic sequence is expressed as Tn = $a+(n-1)d$

a is the first term of the sequence = 100

d is the common difference = 80-100 = 60-80 = 40-60 = 20

n is the number of terms

Substituting the given values in the formula we have;

Tn = $100+(n-1)*-20$

$Tn = 100+(-20n+20)\\Tn = 100-20n+20\\Tn = 120-20n$

The amount of money that remained from Takuya's present during month n is expressed as $Tn = 120-20n$ and the sequence formed is an ARITHMETIC SEQUENCE