General Term of Arithmetic Progression (A.P.)

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Series:

  • If a1, a2, a3, ….., an. is a sequence, then the expression a1 + a+ a3 + …..+ an is called series.

Progression:

  • It is not the case every time that the terms of sequence follow certain pattern so that a specific formula for the nth term of the sequence can be specified explicitly.
  • If the terms of sequence follow certain pattern so that a specific formula for the nth term of the sequence can be specified explicitly, then the sequence is called the progression.

Arithmetic Progression

  • A sequence (tn) is said to be an arithmetic progression (A.P.) if tn + 1 – tn = constant for all n ∈ N
  • The constant difference is called the common difference of the A.P. and is denoted by letter ‘d’. The first term is denoted by the letter ‘a’.

General Term or nth Term of and Sum of an Arithmetic Progression:

  • If ‘a’ is the first term and ‘d’ is the common difference of an A.P. then the nth term is given by

tn = a + (n – 1)d

  • If ‘a’ is the first term and ‘d’ is the common difference of an A.P. then the sum of the first n terms is given by

Stn = n/2(2a + (n – 1)d) = n/2(First term + Last term)

Type – IA: To Find tfor Arithmetic Progression (A.P.).

Algorithm:

  1. Write the first term = a, common difference = d
  2. the nth term of an A.P. is given by tn = a + (n – 1)d
  3. Substitute value of a and d in equation given in step 2
  4. Simplify R.H.S. to get answer.

Example – 01:

  • Find nth term of an arithmetic progression (A.P.) 32, 28, 24, 20, …..
  • Solution:

Given A.P. is 32, 28, 24, 20, …..

First term = a = 32, common difference = d = 28 – 32 = – 4



the nth term of an A.P. is given by tn = a + (n – 1)d

∴  tn = 32 + (n – 1)(-4) = 32 – 4n + 4 = 36 – 4n

Ans: The nth term of the A.P. is (36 – 4n)

Example – 02:

  • Find nth term of an arithmetic progression (A.P.) 4, 9, 14, 19, ……..
  • Solution:

Given A.P. is 4, 9, 14, 19, ……..



First term = a = 4, common difference = d = 9 – 4 = 5

the nth term of an A.P. is given by tn = a + (n – 1)d

 ∴  tn = 4 + (n – 1)(5) = 4 + 5n – 5 = 5n – 1

Ans: The nth term of the A.P. is (5n – 1)

Example – 03:

  • Find tn of an arithmetic progression (A.P.) 4, 14/3, 16/3, 6, ……..
  • Solution:

Given A.P. is 4, 14/3, 16/3, 6, ……..



First term = a = 4, common difference = d = 14/3 – 4 = (14 – 12)/3 = 2/3

the nth term of an A.P. is given by tn = a + (n – 1)d

∴   tn = 4 + (n – 1)(2/3) = 4 + 2n/3 – 2/3 = 2n/3 + 10/3 = (2n+ 10)/3

Ans: The nth term of the A.P. is (2n+ 10)/3

Type – IB: To Find Indicated Term of Arithmetic Progression (A.P.).

Algorithm:

  1. Write the first term = a, common difference = d
  2. the nth term of an A.P. is given by tn = a + (n – 1)d
  3. Substitute value of a and d in equation given in step 2
  4. Simplify R.H.S. to get tn.
  5. Substitute value of n the equation obtained in step 4 and simplify to get answer.

Example – 04:

  • Find 24th term of an arithmetic progression (A.P.) 5, 8, 11, 14, ……....
  • Solution:

Given A.P. is 5, 8, 11, 14, ……..



First term = a = 5, common difference = d = 8 – 5 = 3

the nth term of an A.P. is given by tn = a + (n – 1)d

∴   tn = 5 + (n – 1)(3) = 5 + 3n – 3 = 3n + 2

∴  t24 =  3 x 24 + 2 = 72 + 2 = 74

Ans: The 24th term of the A.P. is 74



Example – 05:

  • Find 15th term of an arithmetic progression (A.P.) 21, 16, 11, 6, ……..
  • Solution:

Given A.P. is 21, 16, 11, 6, ……..

First term = a = 21, common difference = d = 16 – 21 = -5

the nth term of an A.P. is given by tn = a + (n – 1)d

 ∴  tn = 21 + (n – 1)(-5) = 21 – 5n + 5 = 26 – 5n



∴  t15 =  26 – 5 x 15 = 26 – 75 = – 49

Ans: The 15th term of the A.P. is – 49

Example – 06:

  • The 10th term of an arithmetic progression (A.P.) is 1 and 20th term is – 29. Find the 3rd term.
  • Solution:

The nth term of an A.P. is given by tn = a + (n – 1)d

Given t10 = 1 and t20 = – 29

∴  t10 = a + (10 – 1)d = 1

∴  a + 9d = 1  ………… (1)



∴  t20 = a + (20 – 1)d = – 29

∴  a + 19d = – 29  ………… (2)

Subtracting equation (1) from (2) we get

10 d = -30



∴  d = – 3

Substituting in equation (1) we get

a + 9(-3) = 1

∴  a – 27 = 1

∴  a = 28

Now,  tn = a + (n – 1)d



∴   t3 = 28 + (3 – 1)(-3) = 28 – 6 = 22

Ans: The 3rd term of the A.P. is 22

Example – 07:

  • The 7th term of an arithmetic progression (A.P.) is 30 and 10th term is 21. Find the 4th term.
  • Solution:

The nth term of an A.P. is given by tn = a + (n – 1)d

Given t7 = 30 and t10 = 21

∴  t7 = a + (7 – 1)d = 30

∴  a + 6d = 30  ………… (1)

∴  t10 = a + (10 – 1)d = 21

∴  a + 9d = 21  ………… (2)

Subtracting equation (1) from (2) we get

3 d = -9

∴  d = – 3

Substituting in equation (1) we get

a + 6(-3) = 30

∴  a  – 18 = 30

∴  a =  48

Now,  tn = a + (n – 1)d

∴   t4 = 48 + (4 – 1)(-3) = 48 – 9 = 39

Ans: The 4th term of the A.P. is 39

Example – 08:

  • The 3rd term of an arithmetic progression (A.P.) is – 11 and 9th term is – 35. Find the nth term.
  • Solution:

The nth term of an A.P. is given by tn = a + (n – 1)d

Given t3 = – 11 and t9 = – 35

∴  t3 = a + (3 – 1)d = – 11

∴  a + 2d = – 11  ………… (1)

∴  t9 = a + (9 – 1)d = – 35

∴  a + 8d = – 35  ………… (2)

Subtracting equation (1) from (2) we get

6 d = – 24

∴  d = – 4

Substituting in equation (1) we get

a + 2(-4) = – 11

∴  a  – 8 = – 11

∴  a =  = – 3

Now,  tn = a + (n – 1)d

∴    tn = – 3 + (n – 1)(- 4) = – 3 – 4n + 4 = 1 – 4n

Ans: The nth term of the A.P. is (1 – 4n)

Example – 09:

  • Which term of the AP, : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer.
  • Solution:

Given A.P. is 21, 18, 15, . . .

First term = a = 21, common difference = d = 18 – 21 = – 3

The nth term of an A.P. is given by tn = a + (n – 1)d

Given a + (n – 1)d = – 81

∴  21 + (n – 1)(-3) = – 81

∴  (n – 1)(-3) = – 102

∴  n – 1 = 34

∴  n = 35

Thus – 81 is 35th term

Given a + (n – 1)d = 0

∴  21 + (n – 1)(-3) = 0

∴  (n – 1)(-3) = – 21

∴  n – 1 = 7

∴  n = 8

Thus 0 is 8th term

Example – 10:

  • Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .
  • Solution:

Given list. is 5, 11, 17, 23, . . .

First term = a = 5, common difference = d = 11 – 5 = 17 -11 = 23 – 17 = 6

The nth term of an A.P. is given by tn = a + (n – 1)d

Given a + (n – 1)d = 301

∴  5 + (n – 1)(6) = 301

∴  (n – 1)(6) =  296

∴  n – 1 = 296/6 = 148/3

∴  n = 148/3 + 1 = 153/3

But n should be a positive integer but in this case it is fraction

Hence 301 is not a term of the list.

Example – 11:

  • Determine the AP whose 3rd term is 5 and the 7th term is 9.
  • Solution:

The nth term of an A.P. is given by tn = a + (n – 1)d

Given t3 = 5 and t7 = 9

∴  t3 = a + (3 – 1)d = 5

∴  a + 2d = 5  ………… (1)

∴  t7 = a + (7 – 1)d = 9

∴  a + 6d = 9  ………… (2)

Subtracting equation (1) from (2) we get

4d = 4

∴  d = 1

Substituting in equation (1) we get

a + 2(1) = 5

∴   a =  3

Thus the A.P. is 3, 4, 5, 6, ….

Ans: The nth term of the A.P. is (1 – 4n)

Example – 12:

  • If for a sequence (tn), Sn = 4n2 – 3n, find tn and show that the sequence is an A.P.
  • Solution:

Given Sn = 4n2 – 3n  ……….. (1)

Sn – 1 = 4(n – 1)2 – 3(n – 1) = 4(n2 – 2n + 1) – 3n + 3 = 4n2 – 8n + 4 – 3n + 3

Sn – 1 =  4n2 – 11n + 7   ……….. (2)

Subtracting equation (2) from (1)

tn = Sn – Sn – 1 = (4n2 – 3n) – (4n2 – 11n + 7)

tn = 4n2 – 3n – 4n2 + 11n – 7 = 8n – 7

The nth term of an A.P. is tn = 8n – 7  ……….. (3)

∴  tn + 1 = 8(n + 1) – 7 = 8n + 8 – 7

∴  tn + 1 = 8n + 1  ……….. (4)

Subtracting equation (3) from (4) we get

tn + 1 –  tn  = (8n + 1) – (8n – 7) = 8n + 1 -8n + 7 = 8

Thus the common difference is d = 8 which is the constant term

Ans: The sequence (tn) is an A.P.

Example – 13:

  • If for a sequence (tn), Sn = 2n2 + 5n, find tn and show that the sequence is an A.P.
  • Solution:

Given Sn = 2n2 + 5n  ……….. (1)

Sn – 1 = 2(n – 1)2 + 5(n – 1) = 2(n2 – 2n + 1) + 5n – 5 = 2n2 – 4n + 2 + 5n – 5

Sn – 1 =  2n2 + n – 3   ……….. (2)

Subtracting equation (2) from (1)

tn = Sn – Sn – 1 = (2n2 + 5n) – (2n2 + n – 3)

tn = 2n2 + 5n – 2n2 – n + 3 = 4n + 3

The nth term of an A.P. is tn = 4n + 3  ……….. (3)

∴  tn + 1 = 4(n + 1) + 3 = 4n + 4 + 3

∴  tn + 1 = 4n + 7  ……….. (4)

Subtracting equation (3) from (4) we get

tn + 1 –  tn  = (4n + 7) – (4n + 3) = 4n + 74n – 3 = 4

Thus the common difference is d = 4 which is the constant term

Ans: The sequence (tn) is an A.P.

Science > Mathematics > Sequence and SeriesYou are Here
Physics Chemistry  Biology  Mathematics

2 Comments

  1. Error in first example of AP series please check them when anyone see that in starting that’s confused the viewer ..
    Thank you

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