Arithmetic Progression

A sequence of numbers is called an Arithmetic progression if the difference between any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by adding a fixed number to the previous number in the series. For example, 2, 4, 6, 8, 10 is an AP because difference between any two consecutive terms in the series (common difference) is same (4 – 2 = 6 – 4 = 8 – 6 = 10 – 8 = 2).

Fact about Arithmetic Progression :

1. Initial term: In an arithmetic progression, the first number in the series is called the initial term.
2. Common difference: The value by which consecutive terms increase or decrease is called the common difference.

The behavior of the arithmetic progression depends on the common difference d. If the common difference is positive then the members (terms) will grow towards positive infinity or negative, then the members (terms) will grow towards negative infinity.

Formula of Arithmetic Progression:

• The general form of an Arithmetic Progression is a, a + d, a + 2d, a + 3d and so on. Thus nth term of an AP series is T_n = a + (n - 1) d, where T_n = n^th term and a = first term. Here d = common difference = T_n - T_n-1.
• Sum of first n terms of an AP: S =(n/2)[2a + (n- 1)d]
• The sum of n terms is also equal to the formula S_n=n(a+l)/2, where l is the last term.
• T_n = S_n - S_n-1 , where T_n = n^th term
• When three quantities are in AP, the middle one is called as the arithmetic mean of the other two. If a, b and c are three terms in AP   then, b = (a+c)/2
• Arithmetic Progression formula:  n [2a + (n–1) d]/2 = n [a+ a + (n–1) d]/2 =n [a + a_n]/2
  Illustration 1: 11th term of an A.P is 5 and 5th term of an A.P is 11, then what is 16th term?
  Given, a₁₁=5, a₅ = 11
  a + 10d =5—–(1)
  a + 4d= 11—–(2)
  Subtract (2) from (1) from gives,
  6d =−6
  d= −1
  Putting d = -1 in ——(1), a – 10 = 5
  a= 10+5= 15
  a₁₆ = a+15d
  a₁₆ = 15+5(−1) =0
  Note: If mth term of an A.P is n and nth term of an Arithmetic Progression is m, then the common difference of the Arithmetic Progression is -1 and (m+n) th term is zero.

Arithmetic Mean
If number ‘c’ can be inserted between 2 numbers a and b such that a, c, b forms an Arithmetic Progression, then c is called the Arithmetic Mean of a and b
c–a = b–c
2c= a+b
c = (a+b)/2
Illustration 2: Which are the four numbers that can be inserted between 10 and 25 so that the resulting sequence forms an A.P?
Let 10,M1,M2,M3,M4, 25  be the resulting sequence.
Common difference, d = (25–10)/ (4+1) =15/5 = 3
M₁ = 10+3= 13
M₂ = 10+2*3 = 16
M₃= 10+3*3 = 19
M₄ = 10+4*3 =22
10, 13, 16, 19, 22, 25 is an Arithmetic Progression with common difference 3.

Some General Series

• Sum of first n natural numbers =
• Sum of squares of first n natural numbers =
• Sum of cubes of first n natural numbers =

Some examples and solutions
Example 1: An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
A. 25
B. 64
C. 50
D. 27
Explanation: 12 = a + 2d
106 = a + 49d
So, 106-12 = 47d
Or, 94 = 47d
So, d = 2
Hence, a = 8
And, n₂₉ = 8 + 28*2 = 64.

Example 2: What is the sum of all positive integers up to 1000, which are divisible by 5 and are not divisible by 2 ?
A. 10,050
B. 5050
C. 5000
D. 50,000

Explanation: The positive integers, which are exactly divisible by five, are 5, 10, 15, ..., 1000
Out of these 10, 20, 30,,.... 1000 are divisible by two.
Therefore, we will have to find the sum of all the positive integers 5, 15, 25, ...., 995
If n is the no. of terms in it, then sequence is
995 = 5 + 10(n - 1) ⇒ 990 = 10n – 10
1000 = 10n
Thus, n = 100.
Thus the sum of arithmetic progression series = (n/2) (a + l) = (100/2) (5 + 995) = 50000.

Example 3: The sum of the 3 numbers in A.P is 21 & the product of the first & third number of the sequence is 45. What are the 3 numbers ?
A. 5, 7, and 9
B. 9, 7, and 5
C. 3, 7, and 11
D. Both (A) and (B)

Explanation: Let the numbers be a - d, a, a + d
Then a - d + a + a + d = 21
3a = 21
a = 7
and (a - d)(a + d) = 45
a² – d² = 45
d² = 4
d = +2
Hence, the numbers are 5, 7 & 9 when d = 2 and 9, 7 & 5 when d = -2. In both the given cases numbers are the same.

Example 4: The sum of third and ninth term of an A.P is 8. Find the sum of the first 11 terms of the progression.
A. 44
B. 22
C. 19
D. None of the above

Explanation: The third term t₃ = a + 2d
The ninth term t₉ = a + 8d
t₃ + t₉ = 2a + 10d = 8
Sum of 1st 11 terms of an AP is given by
S₁₁ = 11/2 [2a +10d]
S₁₁ = 11/2 * 8 =44

Example 5: What is the sum of the following series? -64, -66, -68, ..... , -100
A. -1458
B. -1558
C. -1568
D. -1664

Explanation: The sum of any set of numbers = Average of the numbers * number of terms
1. Step 1: Compute Average
Average of terms of an arithmetic sequence = [first term + last term/2]
Average of this sequence = [−64 −100]/ 2 = -82
2. Step 2: Compute Sum:
Sum = Average * number of terms
We have computed the no. of terms in the text book approach. Number of terms = 19.
Therefore, sum = (-82) * 19 = -1558

or 
The series is an A.P series with common difference d = -66-(-64 ) = -2 

First term = -64 and last term an = -100

an = a+(n-1)d

-100 = -64 + (n-1) . (-2) 

-100 + 64 = (n-1) . (-2) 

-36 / -2 = (n-1) 

18 = (n-1) 

n = 18 + 1 = 19

avg = -82 ( same as in above solution )

=> sum = 19 x (-82) 

-1558