This lesson will present you the best way to clear up quite a lot of arithmetic sequence phrase issues.

Instance #1:

Suppose that you simply and different college students in your college take part in a fundraising occasion that’s attempting to boost cash for “underprivileged” youngsters . The college begins with $2000 in donations. Then, every scholar should increase not less than $45 in pledges. What number of college students participated if the minimal amount of cash raised is 6050?

Answer

To resolve this downside, we’d like the arithmetic sequence formulation proven beneath.

a_n = a₁ + (n – 1)d

a₁ = starting quantity  = 2000

a_n = remaining quantity  = 6050

d = amount of cash raised by every scholar

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 college students participated within the fundraising occasion.

Instance #2:

The second time period of an arithmetic sequence is -3 and the sixth time period of the arithmetic sequence is 9. Discover the thirtieth time period of the arithmetic sequence.

Answer

Discover the second time period

a_n = a₁ + (n – 1)d

a₂ = a₁ + (2 – 1)d

a₂ = a₁ + d

For the reason that second time period is -3,  -3 = a₁ + d

Discover the sixth time period

a_n = a₁ + (n – 1)d

a₆ = a₁ + (6 – 1)d

a₆ = a₁ + 5d

For the reason that sixth time period is 9,  9 = a₁ + 5d

We now have a system of equations to resolve

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

Difficult arithmetic sequence phrase issues

Instance #3

Buses in your route run each 7 minutes from 6:30 A.M. to 10:00 A.M. You get to the bus cease at 7:56 A.M. How lengthy will you need to look forward to a bus?

Answer

First, you could discover out primarily based on the time you arrived which bus you missed (if any) or which bus is coming if you happen to missed the final one.

a_n = a₁ + (n – 1)d

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

a_n = ready time for the nth bus you want, which can also be equal to the time you arrive on the station.

a_n = 7.56 A.M. – 6:30 A.M. = 86 minutes

d = ready time earlier than the following bus

n = the bus that you could look forward to = ?

86 = 0 + (n – 1)7

86 = 7n – 7

86 + 7 = 7n – 7 + 7

93 = 7n

n = 93 / 3 = 13.285

Discover that 13 < 13.285 < 14

Which means that you arrived after the thirteenth bus had left and earlier than the 14th bus made it to the bus cease. You need to then look forward to the 14th bus!

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

a₁₄ = 0 + (14 – 1)7

a₁₄ = (13)7 = 91 minutes

You arrived 86 minutes after the primary bus and the following bus is coming 91 minutes after the primary bus. Subsequently, you should wait 91 – 86 minutes or 5 minutes.

Instance #4

The arithmetic imply of two phrases in an arithmetic sequence is 56.

1) Discover the opposite time period if one time period is 72

2) Discover the one hundredth time period if the primary time period is 8.

Answer

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

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

15 or the time period between 9 and 21 is the arithmetic imply of 9 and 21

21 or the time period between 15 and 27 is the arithmetic imply of 15 and 27

Normally, if a₁, a₂, a₃, a₄, a₅, … is an arithmetic sequence, a₂ for instance, is the arithmetic imply of a₁ and a₃

Now, you’re prepared to resolve the issue. 

(a₁ + a₃) / 2 = 56

Multiply either side by 2

a₁ + a₃ = 112

For the reason that different time period is 72, we get 72 + a₃ = 112

a₃ = 112 – 72 = 40

Subsequently, the sequence has these phrases: 8, … , 40, 56, 72, …

a_n = 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