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Big Ideas Math Integrated I, 2016

BI

Big Ideas Math Integrated I, 2016 View details

6. Arithemetic Sequences

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Exercise 56 Page 206

The times at which a train stops at the station form an arithmetic sequence with first term $6 : 00$ and common difference $0 : 12 .$

$7$ minutes

Practice makes perfect

Since a train stops at the station every $12$ minutes, the times at which a train stops at the station form an arithmetic sequence with common difference $0 : 12 .$ Also, because the first train stops at $6 : 00 AM,$ the first term of the sequence is $6 : 00 .$ Let's write some terms of the sequence.

$a_{1}$	$6 : 00$
$a_{2}$	$6 : 00 + 0 : 12 = 6 : 12$
$a_{3}$	$6 : 12 + 0 : 12 = 6 : 24$
$a_{4}$	$6 : 24 + 0 : 12 = 6 : 36$
$a_{5}$	$6 : 36 + 0 : 12 = 6 : 48$
$a_{6}$	$6 : 48 + 0 : 12 = 7 : 00$
$a_{7}$	$7 : 00 + 0 : 12 = 7 : 12$
$a_{8}$	$7 : 12 + 0 : 12 = 7 : 24$
$a_{9}$	$7 : 24 + 0 : 12 = 7 : 36$

We see in our table that the next train after

7 : 29 AM

stops at

7 : 36 AM .

Let's find out how many minutes we need to wait.

7 : 36 AM - 7 : 29 AM = 7 minutes

Therefore, we need to wait

7

minutes.

Extra

Note that, in this case, we wrote some terms of the arithmetic sequence instead of using its formula. The reason is that it could be difficult to multiply and add times instead of numbers. However, if special attention is paid and calculations are done very carefully, it is possible to find the answer by using the explicit formula.

Subchapter links

Communicate Your Answer p.199

Monitoring Progress p.200-203

Exercises p.204-206