Arithmetic sequence word problems

This lesson will show you how to solve a variety of arithmetic sequence word problems.

Example #1:

Suppose that you and other students in your school participate in a fundraising event that is trying to raise money for "underprivileged" children . The school starts with $2000 in donations. Then, each student must raise at least $45 in pledges. How many students participated if the minimum amount of money raised is 6050?

Solution

To solve this problem, we need the arithmetic sequence formula shown below.

an = a1 + (n - 1)d

a1 = beginning amount  = 2000

an = final amount  = 6050

d = amount of money raised by each student

6050 = 2000 + (n - 1) × 45

6050 - 2000 = 2000 - 2000 + (n - 1) × 45

4050  = (n - 1) × 45

4050 = 45n - 45 

4050 + 45 = 45n - 45 + 45

4095 = 45n

n = 4095 / 45 = 91

91 students participated in the fundraising event.

Example #2:

The second term of an arithmetic sequence is -3 and the sixth term of the arithmetic sequence is 9. Find the 30th term of the arithmetic sequence.

Solution

Find the second term

an = a1 + (n - 1)d

a2 = a1 + (2 - 1)d

a2 = a1 + d

Since the second term is -3,  -3 = a1 + d

Find the sixth term

an = a1 + (n - 1)d

a6 = a1 + (6 - 1)d

a6 = a1 + 5d

Since the sixth term is 9,  9 = a1 + 5d

We now have a system of equations to solve

a1 + 5d = 9 (equation 1)

a1 + d = -3 (equation 2)

Do equation 1 minus equation 2

a1 - a1 + 5d - d = 9 - -3

4d = 9 + 3

4d = 12

d = 12 / 4 = 3

a1 + d = -3

a1 + 3 = -3

a1 = -3 - 3

a1 = -6

a30 = a1 + (30 - 1)3

a30 = -6 + (29)3

a30 = -6 + 87

a30 = 81

Challenging arithmetic sequence word problems


Example #3

Buses on your route run every 7 minutes from 6:30 A.M. to 10:00 A.M. You get to the bus stop at 7:56 A.M. How long will you have to wait for a bus?

Solution

First, you need to find out based on the time you arrived which bus you missed (if any) or which bus is coming if you missed the last one.

Important observation

a1 = waiting time for first bus = 0 minute

a2 = waiting time for second bus = 0 + 7 minutes 

a3 = waiting time for third bus  =  0 + 2 × 7 minutes 

an = a1 + (n - 1)d

a1 = first bus at 6:30 = 0 minute waiting time

an = waiting time for the nth bus you need, which is also equal to the time you arrive at the station.

an = 7.56 A.M. - 6:30 A.M. = 86 minutes

d = waiting time before the next bus

n = the bus that you need to wait for = ?

86 = 0 + (n - 1)7

86 = 7n - 7

86 + 7 = 7n - 7 + 7

93 = 7n

n = 93 / 3 = 13.285

Notice that 13 < 13.285 < 14

This means that you arrived after the 13th bus had left and before the 14th bus made it to the bus stop. You must then wait for the 14th bus!

Now, find out what time the 14th bus is coming.

a14 = 0 + (14 - 1)7

a14 = (13)7 = 91 minutes

You arrived 86 minutes after the first bus and the next bus is coming 91 minutes after the first bus. Therefore, you must wait 91 - 86 minutes or 5 minutes.

Example #4

The arithmetic mean of two terms in an arithmetic sequence is 56.

1) Find the other term if one term is 72

2) Find the 100th term if the first term is 8.

Solution

The biggest trick here is to know what the "arithmetic mean of two terms in an arithmetic sequence" is.

Suppose 9, 15, 21, 27, ... is an arithmetic sequence.

Notice that (9 + 21) / 2 = 30 / 2 = 15 and (15 + 27) / 2 = 42 / 2 = 21

15 or the term between 9 and 21 is the arithmetic mean of 9 and 21

21 or the term between 15 and 27 is the arithmetic mean of 15 and 27

In general, if a1, a2, a3, a4, a5, ... is an arithmetic sequence, a2 for example, is the arithmetic mean of a1 and a3

Now, you are ready to solve the problem. 

(a1 + a3) / 2 = 56

Multiply both sides by 2

a1 + a3 = 112

Since the other term is 72, we get 72 + a3 = 112

a3 = 112 - 72 = 40

Therefore, the sequence has these terms: 8, ... , 40, 56, 72, ...

an = a1 + (n - 1)d

a1 = 8

n = 100

d = 16 since 56 - 40 = 16 and 72 - 56  = 16

a100 = ?

a100 = 8 + (100 - 1)16

a100 = 8 + (99)16

a100 = 8 + 1584

a100 = 1592

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