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Arithmetic Sequences Simplified Revision Notes

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4.3.1 Arithmetic Sequences

An arithmetic sequence (or arithmetic progression) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is known as the common difference.


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Examples:

  • 1,3,5,7,1, 3, 5, 7, \ldots ✔️
  • 6,4,2,0,6, 4, 2, 0, \ldots ✔️
  • 1,2,4,8,16,1, 2, 4, 8, 16, \ldots ❌ The first term of an arithmetic sequence is denoted by 'aa', the common difference by 'dd', and the term number by 'nn'. The nthn^{th} term is denoted by unu_n.

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Example:

  • For the sequence 5,9,13,17,5, 9, 13, 17, \ldots:
  • a=5a = 5 (first term)
  • d=4d = 4 (common difference)
  • u3=13u_3 = 13 (third term), etc.

Key Concepts:

  1. General Form:
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An arithmetic sequence can be written as:

a,a+d,a+2d,a+3d,a, \, a + d, \, a + 2d, \, a + 3d, \, \dots

  1. nth Term of an Arithmetic Sequence: The nn th term (or general term) of an arithmetic sequence is given by the formula:

an=a+(n1)da_n = a + (n - 1)d

  • ana_n is the nn th term.
  • aa is the first term.
  • dd is the common difference.
  • nn is the position of the term in the sequence.
infoNote

Example: Consider the sequence 3,7,11,15,3, 7, 11, 15, \dots :

  • Here, a=3a = 3 and d=4d = 4 .
  • To find the 5th5th term ( a5a_5 ):

a5=3+(51)×4=3+16=19a_5 = 3 + (5 - 1) \times 4 = 3 + 16 = 19

So, the 5th5th term is 1919.

  1. Sum of the First nn Terms (Arithmetic Series): The sum of the first nn terms of an arithmetic sequence (also called an arithmetic series) is given by:

Sn=n2(2a+(n1)d)S_n = \frac{n}{2} \left( 2a + (n - 1)d \right)

  • SnS_n is the sum of the first nn terms.
  • nn is the number of terms.
  • aa is the first term.
  • dd is the common difference. Alternatively, it can also be written as:

Sn=n2(a+an)S_n = \frac{n}{2} \left( a + a_n \right)

  • Where ana_n is the nn th term.
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Example: Using the previous sequence 3,7,11,15,3, 7, 11, 15, \dots , find the sum of the first 5 terms:

  • First term a=3a = 3 .
  • Common difference d=4d = 4 .
  • Number of terms n=5n = 5 .
  • The 5th5th term a5=19a_5 = 19 . Now, using the sum formula:

S5=52(3+19)=52×22=5×11=55S_5 = \frac{5}{2} \left( 3 + 19 \right) = \frac{5}{2} \times 22 = 5 \times 11 = 55

So, the sum of the first 55 terms is 5555.


Example Problem:

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Problem: Given an arithmetic sequence with the first term a=2a = 2 and the common difference d=5d = 5 , find the 10th term and the sum of the first 10 terms.

  1. Finding the 10th term:

a10=2+(101)×5=2+9×5=2+45=4a_{10} = 2 + (10 - 1) \times 5 = 2 + 9 \times 5 = 2 + 45 = 47

So, the 10th10th term is 4747. 4. Finding the sum of the first 10 terms:

S10=102×(2+47)=5×49=245S_{10} = \frac{10}{2} \times (2 + 47) = 5 \times 49 = 245

So, the sum of the first 1010 terms is 245245.

Summary:

infoNote
  • Arithmetic Sequence: A sequence where each term is found by adding a constant difference to the previous term.
  • nth Term: an=a+(n1)da_n = a + (n - 1)d .
  • Sum of First nn Terms: Sn=n2(2a+(n1)d)S_n = \frac{n}{2} \left( 2a + (n - 1)d \right) or Sn=n2(a+an)S_n = \frac{n}{2} \left( a + a_n \right) .
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