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# Arithmetic Progression

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#### Arithmetic Progression - Lesson Summary

A collection of numbers arranged in a definite order according to some definite rule or pattern is called a sequence.

Each number in the sequence is called a term.
Term in the n th position is called the nth term, denoted by a n.
A sequence is said to be in arithmetic progression, if every term (except the first) is obtained by adding a fixed number to its preceding term.

The fixed number is called the common difference, and is obtained by subtracting any term from its succeeding term.

d = a n â€“ a n - 1

General form of an AP

a, a+d, a+2d, a+3d, a+4d, â€¦ where a is the first term and d is the common difference.

Here, (a + d) is the second term, (a + 2d) is the third term, (a + 3d) is the fourth term, and so on.

Likewise, the n th term, which is known as the â€˜general term,â€™ is given by a 1n = a + n(1)d.

With the help of this general term or n th term, any required term can be found in the sequence without actually finding the preceding terms.

For an AP: a, a+d, a+2d, a+3d, a+4d, â€¦

S n = n 2 [2a + (n - 1)d]

or S n = n 2 [a + l] , where l = a + n(1)d.

Properties of Arithmetic Progression

i. If a constant is added to each term of an AP, then the resulting sequence is also an AP.
ii. If a constant is subtracted from each term of an AP, then the resulting sequence is also an AP.
iii. If each term of an AP is multiplied by a constant, then the resulting sequence is also an AP.
iv.If each term of an AP is divided by a non-zero constant, then the resulting sequence is also an AP.
v. If {a 1,a 2,a 3,.....}  and {b 1,b 2,b 3,....}  are two APâ€™s, then a 1 + b 1,a 2 + b 2,a 3 + b 3,....  are also in AP.

If {a 1,a 2,a 3,.....}  and {b 1,b 2,b 3,....}  are two APâ€™s, then a 1 - b 1,a 2 - b 2,a 3 - b 3,....  are also in AP, if and only if the common difference of the two sequences is not the same.

If a constant is added to each term of an arithmetic progression, then the resulting sequence is also an arithmetic progression.

Example: Let 2, 4, 6, 8, â€¦â€¦ be an AP.

Add 3 to each term of the given sequence.

The resultant sequence is 5, 7, 9, 11, â€¦

Common difference = 7 â€“ 5 = 9 â€“ 7 = 2.

Similarly, if a constant is subtracted from each term of an AP, the resulting sequence is also an AP.
if each term of an arithmetic progression is multiplied by a constant, then the resulting sequence is also an arithmetic progression.

Example: Let 5, 8, 11, 14â€¦ be an AP.

Common difference (d) = 3

Multiply each term of the sequence by 2.

The resultant sequence is 10, 16, 22, 28â€¦, which is also an AP.

Common difference (d 1) = 6

The common difference of the resultant sequence is also a multiple of the number with which the sequence is multiplied.

d 1 =2 d

Similarly, if each term of an arithmetic progression is divided by a non-zero constant, then the resulting sequence is also an arithmetic progression.

If {a 1,a 2,a 3,.....}  and {b 1,b 2,b 3,....}  are two APâ€™s, then a 1 + b 1,a 2 + b 2,a 3 + b 3,.... are also in AP.

Example: Let 2,4,6,8... and 3,5,7,9,â€¦ be two APâ€™s.

Then (2+3), (4+5), (6+7), (8+9)â€¦ â‡’ 5, 9, 13, 17, â€¦ is also an AP.

Common difference = 9 â€“ 5 = 13 â€“ 9 = 4 = 2 + 2

Here, the common difference of the resultant sequence is the sum of the common differences of the two given APâ€™s.

Similarly, if the corresponding terms of two arithmetic progressions are subtracted, then the resulting sequence is also an arithmetic progression, provided that the common difference of the two given sequences is not the same.

Result: If the n th term of any sequence is a linear expression in n, then the sequence is an AP.

Proof:
Let t n = an+b, where a and b are real constants and  n=1, 2, 3, â€¦

For n=1, we have t 1 = a.1 + b = a+b

For n=2, we have t 2 = a.2 + b = 2a+b

For n=3, we have t 3 = a.3 + b = 3a+b

t n-1 = a.(n - 1) + b

t 2 - t 1 = (2a + b) + (a + b) = a

t 3 - t 2 = (3a + b) + (2a + b) = a

t n â€“ t n - 1 = (an + b) â€“ (a(n â€“ 1) + b) = an + b â€“ an â€“ b = a

Therefore, the common difference of the sequence is a, which is a constant.

Hence, a sequence whose n th term is a linear expression in n forms an AP.

Arithmetic mean

To form an arithmetic progression by inserting the desired number of terms between any two terms:

Given two numbers a and b, we can find a number, c, midway between them such that the sequence, a, c, b, forms an arithmetic progression.

That is, c - a = b - c

or 2c = a + b

or c = a + b 2

Such a number â€˜câ€™ is called the arithmetic mean of the numbers a and b.

The arithmetic mean (AM) between two numbers is a number, which, when placed between them, forms with them an arithmetic progression.

In general, we can insert any number of terms between two given terms such that the resulting sequence is an arithmetic progression.
Let A 1, A 2, A 3,....,A n be the n terms between the two numbers a and b such that a,A 1, A 2, A 3,....,A n,b forms an AP.

The resultant AP has (n + 2) terms, where a 1 = a, a 2 = A 1,a 3 = A 2,...,a n+1 = A n, a n+2 = b  .

Now, a n+2 = b

â‡’ a + n + 2(1)d = b, where d is the common difference of this AP

a + (n+1)d = b

(n+1)d = b - a

d = b - a n + 1

Now, A 1, A 2, A 3,....,A n can all be obtained by adding d successively to a, the first term.

Example:

Insert 4 numbers between 2 and 32 such that the resulting sequence is an arithmetic progression.

Solution: Let A 1, A 2, A 3, and A 4 be the four numbers between 2 and 32 such that 2, A 1, A 2, A 3, A 4, 32 form
an AP.

Number of terms n = 6

32 = 2 + (6 - 1)d   [ âˆµ a 1n = a + n(1)d ]

Thus, A 1 = 2 + d = 2 + 6  = 8

A 2 = 2 + 2d = 2 + (2 x 6)  = 14

A 3 = 2 + 3d = 2 + (3 x 6)  = 20

A 4 = 2 + 4d = 2 + (4 x 6)  = 26

Hence, the four numbers between 2 and 32 are 8, 14, 20 and 26.