Arithmetic Progression

In this article, we shall study series, progression, and arithmetic progression (A.P.)

Series:

If a₁, a₂, a₃, ….., a_n. is a sequence, then the expression a₁ + a₂ + a₃ + …..+ a_n 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 (t_n) is said to be an arithmetic progression (A.P.) if t_{n + 1} – t_n = constant for all n ∈ N. The constant difference is called the common difference of the A.P. and is denoted by the letter ‘d’. The first term is denoted by the letter ‘a’.

General Term or n^th term of an Arithmetic Progression:

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

t_n = a + (n – 1)d

Sum up to n^th term of an Arithmetic Progression

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

S_n = n/2(First term + Last term)

To Find t_n for Arithmetic Progression (A.P.):

Algorithm:

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

Example – 01:

Find n^th 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 n^th term of an A.P. is given by t_n = a + (n – 1)d

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

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

Example – 02:

Find n^th 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 n^th term of an A.P. is given by t_n = a + (n – 1)d

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

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

Example – 03:

Find t_n 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 n^th term of an A.P. is given by t_n = a + (n – 1)d

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

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

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

Algorithm:

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

Example – 04:

Find 24^th 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 n^th term of an A.P. is given by t_n = a + (n – 1)d

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

∴  t₂₄ =  3 x 24 + 2 = 72 + 2 = 74

Ans: The 24^th term of the A.P. is 74

Example – 05:

Find 15^th 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 n^th term of an A.P. is given by t_n = a + (n – 1)d

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

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

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

Example – 06:

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

Solution:

The n^th term of an A.P. is given by t_n = 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,  t_n = a + (n – 1)d

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

Ans: The 3^rd term of the A.P. is 22

Example – 07:

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

Solution:

The n^th term of an A.P. is given by t_n = 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,  t_n = a + (n – 1)d

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

Ans: The 4^th term of the A.P. is 39

Example – 08:

The 3^rd term of an arithmetic progression (A.P.) is – 11 and 9^th term is – 35. Find the n^th term.

Solution:

The n^th term of an A.P. is given by t_n = 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,  t_n = a + (n – 1)d

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

Ans: The n^th 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 a reason for your answer.

Solution:

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

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

The n^th term of an A.P. is given by t_n = 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 35^th term

Given a + (n – 1)d = 0

∴  21 + (n – 1)(-3) = 0

∴  (n – 1)(-3) = – 21

∴  n – 1 = 7

∴  n = 8

Thus 0 is 8^th 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 n^th term of an A.P. is given by t_n = 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 n^th term of an A.P. is given by t_n = 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 n^th term of the A.P. is (1 – 4n)

Example – 12:

If for a sequence (t_n), S_n = 4n² – 3n, find t_n and show that the sequence is an A.P.

Solution:

Given S_n = 4n² – 3n  ……….. (1)

S_{n – 1} = 4(n – 1)² – 3(n – 1) = 4(n² – 2n + 1) – 3n + 3 

S_n = 4n² – 8n + 4 – 3n + 3

S_{n – 1} =  4n² – 11n + 7   ……….. (2)

Subtracting equation (2) from (1)

t_n = S_n – S_{n – 1} = (4n² – 3n) – (4n² – 11n + 7)

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

The n^th term of an A.P. is t_n = 8n – 7  ……….. (3)

∴  t_{n + 1} = 8(n + 1) – 7 = 8n + 8 – 7

∴  t_{n + 1} = 8n + 1  ……….. (4)

Subtracting equation (3) from (4) we get

t_{n + 1} –  t_n  = (8n + 1) – (8n – 7) = 8n + 1 -8n + 7 = 8

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

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

Example – 13:

If for a sequence (t_n), S_n = 2n² + 5n, find t_n and show that the sequence is an A.P.

Solution:

Given S_n = 2n² + 5n  ……….. (1)

S_{n – 1} = 2(n – 1)² + 5(n – 1) = 2(n² – 2n + 1) + 5n – 5

S_n = 2n² – 4n + 2 + 5n – 5

S_{n – 1} =  2n² + n – 3   ……….. (2)

Subtracting equation (2) from (1)

t_n = S_n – S_{n – 1} = (2n² + 5n) – (2n² + n – 3)

t_n = 2n² + 5n – 2n² – n + 3 = 4n + 3

The n^th term of an A.P. is t_n = 4n + 3  ……….. (3)

∴  t_{n + 1} = 4(n + 1) + 3 = 4n + 4 + 3

∴  t_{n + 1} = 4n + 7  ……….. (4)

Subtracting equation (3) from (4) we get

t_{n + 1} –  t_n  = (4n + 7) – (4n + 3) = 4n + 7 – 4n – 3 = 4

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

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