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Sequence and Series

Arithmetic Sequence

Nabil Akbarazzima Fatih

Mathematics

Understanding Arithmetic Sequences

An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is always constant. This difference in an arithmetic sequence is denoted by bb.

Examples of Arithmetic Sequences

Consider the following number sequence:

4,6,8,10,...4, 6, 8, 10, ...

In this sequence, we can see that:

  • The difference between the second and first term: 64=26 - 4 = 2
  • The difference between the third and second term: 86=28 - 6 = 2
  • The difference between the fourth and third term: 108=210 - 8 = 2

Since the difference between consecutive terms is always 2, this sequence is an arithmetic sequence with common difference b=2b = 2.

Common Difference in Arithmetic Sequences

The common difference (bb) in an arithmetic sequence can be calculated by subtracting consecutive terms:

b=U2U1=U3U2=U4U3=...=UnUn1b = U_2 - U_1 = U_3 - U_2 = U_4 - U_3 = ... = U_n - U_{n-1}

Where:

  • UnU_n represents the nn th term
  • Un1U_{n-1} represents the (n1n-1 )th term

Formula for the nth Term

General Formula

To determine the nnth term of an arithmetic sequence, we can use the formula:

Un=a+(n1)bU_n = a + (n-1)b

Where:

  • UnU_n = the nn th term
  • aa = first term
  • nn = term number
  • bb = common difference

Deriving the Formula

If we write the first few terms of an arithmetic sequence:

  • 1st term: U1=aU_1 = a
  • 2nd term: U2=a+bU_2 = a + b
  • 3rd term: U3=a+2bU_3 = a + 2b
  • 4th term: U4=a+3bU_4 = a + 3b
  • 5th term: U5=a+4bU_5 = a + 4b

From this pattern, we can see that the nnth term is:

Un=a+(n1)bU_n = a + (n-1)b

Applications of Arithmetic Sequences

Performing Arts Theater

Consider the number of seats in a performing arts theater:

  • Row 1 = 20 seats
  • Row 2 = 24 seats
  • Row 3 = 28 seats
  • Row 4 = 32 seats
  • Row 5 = 36 seats

To determine the number of seats in a specific row, we need to find the pattern in this data.

Step 1: Finding the common difference between rows

  • Row 2 - Row 1: 2420=424 - 20 = 4
  • Row 3 - Row 2: 2824=428 - 24 = 4
  • Row 4 - Row 3: 3228=432 - 28 = 4
  • Row 5 - Row 4: 3632=436 - 32 = 4

We can see that the difference between the number of seats in consecutive rows is 4. This means the number of seats in this theater forms an arithmetic sequence with:

  • First term (aa) = 20
  • Common difference (bb) = 4

Step 2: Using the formula to find the number of seats in row 15

U15=a+(n1)bU_{15} = a + (n-1)b
U15=20+(151)4U_{15} = 20 + (15-1)4
U15=20+(14)4U_{15} = 20 + (14)4
U15=20+56U_{15} = 20 + 56
U15=76U_{15} = 76

Therefore, there are 76 seats in row 15.

Exercise 1

Given an arithmetic sequence with the 3rd term = 9 and the 6th term = 18. Find the formula for the nth term.

Solution to Exercise 1

To determine the formula for the nth term, we need to find the values of the first term (a)(a) and the common difference (b)(b).

U3=a+2b=9U_3 = a + 2b = 9
U6=a+5b=18U_6 = a + 5b = 18

We eliminate these equations to find the value of b:

U6U3=(a+5b)(a+2b)U_6 - U_3 = (a + 5b) - (a + 2b)
189=3b18 - 9 = 3b
9=3b9 = 3b
b=3b = 3

After finding b, we substitute it into the first equation to find a:

a+2b=9a + 2b = 9
a+2(3)=9a + 2(3) = 9
a+6=9a + 6 = 9
a=96=3a = 9 - 6 = 3

With a = 3 and b = 3, we can formulate the nth term:

Un=a+(n1)bU_n = a + (n-1)b
Un=3+(n1)3U_n = 3 + (n-1)3
Un=3+3n3U_n = 3 + 3n - 3
Un=3nU_n = 3n

Therefore, the formula for the nth term of this sequence is Un=3nU_n = 3n

Exercise 2

Rudi saves money in a bank with a constant monthly increase. If in the 5th month, he saves Rp70,000.00 and in the 9th month, Rudi saves Rp90,000.00.

a. What is the monthly increase in savings amount?

b. How much money did Rudi save for the first time?

Solution to Exercise 2

Rudi's savings form an arithmetic sequence because the monthly increase is constant.

a. Finding the monthly increase in savings

U5=70,000U_5 = 70,000
a+(51)b=70,000a + (5-1)b = 70,000
a+4b=70,000 ... (equation 1)a + 4b = 70,000 \text{ ... (equation 1)}
U9=90,000U_9 = 90,000
a+(91)b=90,000a + (9-1)b = 90,000
a+8b=90,000 ... (equation 2)a + 8b = 90,000 \text{ ... (equation 2)}

Eliminating equations 1 and 2:

4b=20,0004b = 20,000
b=5,000b = 5,000

Therefore, the monthly increase in Rudi's savings is Rp5,000.00.

b. Finding Rudi's initial savings

We have found b = 5,000, now we substitute it into equation 1 to find a:

a+4b=70,000a + 4b = 70,000
a+4(5,000)=70,000a + 4(5,000) = 70,000
a+20,000=70,000a + 20,000 = 70,000
a=70,00020,000a = 70,000 - 20,000
a=50,000a = 50,000

Therefore, Rudi's initial savings was Rp50,000.00.