What Is Arithmetic Progression

What Is Arithmetic Progression

Arithmetic Progression (A.P.)

Arithmetic Progression is defined as a series in which difference between any two consecutive terms is constant throughout the series. This constant difference is called common difference.

A sequence of numbers < t_n > is said to be in arithmetic progression (A.P.) when the difference t_n – t_n–1 is a constant for all n ∈ N. This constant is called the common difference of the A.P. and is usually denoted by the letter d.

If ‘a’ is the first term and ‘d’ the common difference, then an A.P. can be represented as a + (a + d) + (a + 2d) + (a + 3d) + ……

Example : 2, 7, 12, 17, 22, …… is an A.P. whose first term is 2 and common difference 5.

Algorithm to determine whether a sequence is an A.P. or not.

Step I: Obtain a_n (the n^th term of the sequence).

Step II: Replace n by n – 1 in a_n to get a_n–1.

Step III: Calculate a_n – a_n–1.

If a_n – a_n–1 is independent from n, the given sequence is an A.P. otherwise it is not an A.P.

∴ t_n = An + B represents the n^th term of an A.P. with common difference A.

Note: If a,b,c, are in AP ⟺ 2b = a + c

General term of an A.P.

(1) Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. Then its n^th term is a + (n– 1) d i.e., T_n = a + (n– 1) d.

(2) r^th term of an A.P. from the end: Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. having n terms. Then r^th term from the end is (n – r + 1)^th term from the beginning

i.e., r^th term from the end = T_(n-r+1) = a + (n – r)d.

If last term of an A.P. is l then r^th term from end = l – (r – 1)d.

Selection of terms in an A.P.

When the sum is given, the following way is adopted in selecting certain number of terms :

In general, we take a – rd, a – (r – 1)d, ……., a – d, a, a + d, ……, a + (r – 1)d, a + rd, in case we have to take (2r + 1) terms (i.e. odd number of terms) in an A.P.

And, a – (2r – )d, a – (2r – 3)d,……..,  a – d, a + d, ……, a + (2r – 1)d in case we have to take 2r terms in an A.P.

When the sum is not given, then the following way is adopted in selection of terms.

Sum of n terms of an A.P.

The sum of n terms of the series a, (a + d), (a + 2d), (a + 3d), …… {a + (n – 1)d} is given by

Arithmetic mean

If a, A, b are in A.P., then A is called A.M. between a and b.

(1) If a, A₁, A₂, A₃, ….. A_n, b are in A.P., then A₁, A₂, A₃, ….. A_n are called n A.M.’s between a and b.

(2) Insertion of arithmetic means

(i) Single A.M. between a and b :

If a and b are two real numbers then single A.M. between a and b = (frac { a+b }{ 2 })

(ii) n A.M.’s between a and b : If A₁, A₂, A₃, ….. A_n  are n A.M.’s between a and b, then

Properties of A.P.

1. If a₁, a₂, a₃, …… are in A.P. whose common difference is d, then for fixed non-zero number k ∈ R.
   1. a₁ ± k, a₂ ± k, a₃ ± k, .….. will be in A.P., whose common difference will be d.
   2. ka₁, ka₂, ka₃, …… will be in A.P. with common difference = kd.
   3. a₁/k, a₂/k, a₃/k, ……will be in A.P. with common difference = d/k.
2. The sum of terms of an A.P. equidistant from the beginning and the end is constant and is equal to sum of first and last term. i.e. a₁ + a_n = a₂ + a_n−1 = a3 + a_n−2 = …..
3. If number of terms of any A.P. is odd, then sum of the terms is equal to product of middle term and number of terms.
4. If number of terms of any A.P. is even then A.M. of middle two terms is A.M. of first and last term.
5. If the number of terms of an A.P. is odd then its middle term is A.M. of first and last term.
6. If a₁, a₂,…… a_n and b₁, b₂,…… b_n are the two A.P.’s. Then a₁ ± b₁, a₂ ± b₂,.….. a_n ± b_n are also A.P.’s with common difference d₁ ≠ d₂, where d₁ and d₂ are the common difference of the given A.P.’s.
7. Three numbers a, b, c are in A.P. iff 2b = a + c.
8. If T_n, T_n+1, and T_n+2 are three consecutive terms of an A.P., then 2T_n+1 = T_n + T_n+2.
9. If the terms of an A.P. are chosen at regular intervals, then they form an A.P.

How to Find n^th Term in Arithmetic Progression With Examples

Example 1:    Find 26^th term from last of an AP 7, 15, 23……., 767 consits 96 terms.

Solution.    Method: I

r^th term from end is given by

= T_n – (r – 1) d        or

= (n – r + 1)^th term from beginning where n is total no. of terms.

m = 96, n = 26


∴ T₂₆ from last = T_(96-26+1)  from beginning

= T₇₁ from beginning

= a + 70d

= 7 + 70 (8) = 7 + 560 = 567

Method: II

d = 15 – 7 = 8

∴ from last, a = 767 and d = –8

∴ T₂₆ = a + 25d = 767 + 25 (–8)

= 767 – 200

= 567.

Example 2:    If the n^th term of a progression be a linear expression in n, then prove that this progression is an AP.

Solution.    Let the n^th term of a given progression be given by

T_n = an + b, where a and b are constants.

Then, T_n-1 = a(n – 1) + b = [(an + b) – a]

∴ (T_n – T_n-1) = (an + b) – [(an + b) – a] = a,

which is a constant.

Hence, the given progression is an AP.

Example 3:    Write the first three terms in each of the sequences defined by the following –

(i) a_n = 3n + 2                 (ii) a_n = n² + 1

Solution.   (i) We have,

a_n = 3n + 2

Putting n = 1, 2 and 3, we get

a₁ = 3 × 1 + 2 = 3 + 2 = 5,

a₂ = 3 × 2 + 2 = 6 + 2 = 8,

a₃ = 3 × 3 + 2 = 9 + 2 = 11

Thus, the required first three terms of the sequence defined by a_n = 3n + 2 are 5, 8, and 11.

(ii) We have,

a_n = n² + 1

Putting n = 1, 2, and 3 we get

a₁ = 12 + 1 = 1 + 1 = 2

a₂ = 22 + 1 = 4 + 1 = 5

a₃ = 32 + 1 = 9 + 1 = 10

Thus, the first three terms of the sequence defined by a_n = n² + 1 are 2, 5 and 10.

Example 4:    Write the first five terms of the sequence defined by a_n = (–1)^n-1 . 2ⁿ

Solution.    a_n = (–1)^n-1 × 2ⁿ

Putting n = 1, 2, 3, 4, and 5 we get

a₁ = (–1)^1-1 × 2¹ = (–1)⁰ × 2 = 2

a₂ = (–1)^2-1 × 2² = (–1)¹ × 4 = – 4

a₃ = (–1)^3-1 × 2³ = (–1)² × 8 × 8

a₄ = (–1)^4-1 × 2⁴ = (–1)³ × 16 = –16

a₅ = (–1)^5-1 × 2⁵ = (–1)⁴ × 32 = 32

Thus the first five term of the sequence are 2, –4, 8, –16, 32.

Example 5:    The nth term of a sequence is 3n – 2. Is the sequence an A.P. ? If so, find its 10^th term.

Solution.     We have a_n = 3n – 2

Clearly an is a linear expression in n. So, the given sequence is an A.P. with common difference 3.

Putting n = 10, we get

a₁₀ = 3 × 10 – 2 = 28

Example 6:     Find the 12^th, 24^th and n^th term of the A.P. given by 9, 13, 17, 21, 25, ………

Solution.     We have,

a = First term = 9 and, d = Common difference = 4

[∵ 13 – 9 = 4, 17 – 13 = 4, 21 – 7 = 4 etc.]

We know that the nth term of an A.P. with first term a and common difference d is given by a_n = a + (n – 1) d

Therefore,

a₁₂ = a + (12 – 1) d = a + 11d = 9 + 11 × 4 = 53

a₂₄ = a + (24 – 1) d = a + 23 d = 9 + 23 × 4 = 101

and, a_n = a + (n – 1) d = 9 + (n – 1) × 4 = 4n + 5

a₁₂ = 53, a₂₄ = 101 and a_n = 4n + 5

Example 7:     Which term of the sequence –1, 3, 7, 11, ….. , is 95 ?

Solution.    Clearly, the given sequence is an A.P.

We have,

a = first term = –1 and, d = Common difference = 4.

Let 95 be the n^th term of the given A.P. then,

a_n = 95

⇒ a + (n – 1) d = 95

⇒ – 1 + (n – 1) × 4 = 95

⇒ – 1 + 4n – 4 = 95   ⇒    4n – 5 = 95

⇒ 4n = 100 ⇒ n = 25

Thus, 95 is 25^th term of the given sequence.

Example 8:     Which term of the sequence 4, 9 , 14, 19, …… is 124 ?

Solution.    Clearly, the given sequence is an A.P. with first term a = 4 and common difference d = 5.

Let 124 be the n^th term of the given sequence. Then, a_n = 124

a + (n – 1) d = 124

⇒ 4 + (n – 1) × 5 = 124

⇒ n = 25

Hence, 25^th term of the given sequence is 124.

Example 9:    The 10^th term of an A.P. is 52 and 16^th term is 82. Find the 32nd term and the general term.

Solution.    Let a be the first term and d be the common difference of the given A.P. Let the A.P. be

a₁, a₂, a₃, ….. an, ……

It is given that a₁₀ = 52 and a₁₆ = 82

⇒ a + (10 – 1) d = 52 and a + (16 – 1) d = 82

⇒ a + 9d = 52        ….(i)

and, a + 15d = 82         ….(ii)

Subtracting equation (ii) from equation (i), we get

–6d = – 30 ⇒ d = 5

Putting d = 5 in equation (i), we get

a + 45 = 52 ⇒ a = 7

∴ a₃₂ = a + (32 – 1) d = 7 + 31 × 5 = 162

and, an = a + (n – 1) d = 7 (n – 1) × 5 = 5n + 2.

Hence a₃₂ = 162 and an = 5n + 2.

Example 10:    Determine the general term of an A.P. whose 7^th term is –1 and 16^th term 17.

Solution.    Let a be the first term and d be the common difference of the given A.P. Let the A.P. be a1, a₂, a₃, ……. a_n, …….

It is given that a₇ = – 1 and a₁₆ = 17

a + (7 – 1) d = – 1 and, a + (16 – 1) d = 17

⇒ a + 6d = – 1             ….(i)

and, a + 15d = 17        ….(ii)

Subtracting equation (i) from equation (ii), we get

9d = 18 ⇒ d = 2

Putting d = 2 in equation (i), we get

a + 12 = – 1 ⇒ a = – 13

Now, General term = an

= a + (n – 1) d = – 13 + (n – 1) × 2 = 2n – 15

Example 11:    If five times the fifth term of an A.P. is equal to 8 times its eight term, show that its 13^th term is zero.

Solution.    Let a₁, a₂, a₃, ….. , an, …. be the A.P. with its first term = a and common difference = d.

It is given that 5a₅ = 8a₈

⇒ 5(a + 4d) = 8 (a + 7d)

⇒ 5a + 20d = 8a + 56d ⇒ 3a + 36d = 0

⇒ 3(a + 12d) = 0 ⇒ a + 12d = 0

⇒ a + (13 – 1) d = 0 ⇒ a₁₃ = 0

Example 12:    If the m^th term of an A.P. be 1/n and n^th term be 1/m, then show that its (mn)^th term is 1.

Solution.    Let a and d be the first term and common difference respectively of the given A.P. Then,

1/n= m^th term   ⇒   1/n = a + (m – 1) d       ….(i)

1/m = n^th term   ⇒   1/m = a + (n – 1) d        ….(ii)

On subtracting equation (ii) from equation (i), we get

( frac{1}{n}-frac{1}{m}=~left( mn right)d )

( Rightarrow frac{m-n}{mn}=~left( mn right)dRightarrow d=frac{1}{mn} )

( ~text{Putting d}=~frac{1}{mn}text{in equation }left( text{i} right)text{, we get} )

( frac{1}{n}
=a+frac{(m-1)}{mn}Rightarrow a=frac{1}{mn} )

∴  (mn)^th term = a + (mn – 1) d

( =frac{1}{mn}+(mn-1)frac{1}{mn}=1 )

Example 13:    If m times m^th term of an A.P. is equal to n times its n^th term, show that the (m + n) term of the A.P. is zero.

Solution.    Let a be the first term and d be the common difference of the given A.P. Then, m times m^th term = n times n^th term

⇒ ma_m = na_n

⇒ m{a + (m – 1) d} = n {a + (n – 1) d}

⇒ m{a + (m – 1) d} – n{a + (n – 1) d} = 0

⇒ a(m – n) + {m (m – 1) – n(n – 1)} d = 0

⇒ a(m – n) + (m – n) (m + n – 1) d = 0

⇒ (m – n) {a + (m + n – 1) d} = 0

⇒ a + (m + n – 1) d = 0

⇒ a_m+n = 0

Hence, the (m + n)^th term of the given A.P. is zero.

Example 14:    If the pth term of an A.P. is q and the qth term is p, prove that its nth term is (p + q – n).

Solution.     Let a be the first term and d be the common difference of the given A.P. Then,

p^th term = q   ⇒   a + (p – 1) d = q           ….(i)

q^th term = p   ⇒   a + (q – 1) d = p          ….(ii)

Subtracting equation (ii) from equation (i),

we get

(p – q) d = (q – p)   ⇒   d = – 1

Putting d = – 1 in equation (i), we get

a = (p + q – 1)

n^th term = a + (n – 1) d

= (p + q – 1) + (n – 1) × (–1) = (p + q – n)

Example 15:    If p^th, q^th and r^th terms of an A.P. are a, b, c respectively, then show that

(i) a (q – r) + b(r – p) + c(p – q) = 0

(ii) (a – b) r + (b –¬ c) p + (c – a) q = 0

Solution.     Let A be the first term and D be the common difference of the given A.P. Then,

a = p^th term   ⇒   a = A + (p – 1) D          ….(i)

b = q^th term   ⇒   b = A + (q – 1) D         ….(ii)

c = r^th term   ⇒   c = A+ (r – 1) D            ….(iii)

(i)  We have,

a(q – r) + b (r – p) + c (p – q)

= {A + (p – 1) D} (q – r) + {A + (q – 1)} (r – p) + {A + (r – 1) D} (p – q)

[Using equations (i), (ii) and (iii)]

= A {(q – r) + (r – p) + (p – q)} + D {(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)}

= A {(q – r) + (r – p) + (p – q)} + D{(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)}

= A . 0 + D {p (q – r) + q (r – p) + r (p – q) – (q – r) – (r – p) – (p – q)}

= A . 0 + D . 0 = 0

(ii)  On subtracting equation (ii) from equation (i), equation (iii) from equation (ii) and equation (i) from equation (iii), we get

a – b = (p – q) D, (b – c) = (q – r) D and c – a = (r – p) D

∴ (a – b) r + (b – c) p + (c – a) q

= (p – q) Dr + (q – r) Dp + (r – p) Dq

= D {(p – q) r + (q – r) p + (r – p) q}

= D × 0 = 0

Example 16:    Determine the 10th term from the end of the A.P. 4, 9, 14, …….., 254.

Solution.    We have,

l = Last term = 254 and,

d = Common difference = 5,

10^th term from the end = l – (10 – 1) d

= l – 9d = 254 – 9 × 5 = 209.

Example 17:    Four numbers are in A.P. If their sum is 20 and the sum of their square is 120, then find the middle terms.

Solution.    Let the numbers are a – 3d, a – d, a + d, a + 3d

given a – 3d + a – d + a + d + a + 3d = 20

⇒ 4a = 20   ⇒  a = 5

and (a – 3d)² + (a – d)2 + (a + d)² + (a + 3d)² = 120

4a² + 20 d² = 120

4 × 5² + 20 d² = 120

d² = 1   ⇒  d = ±1

Hence numbers are 2, 4, 6, 8

Example 18:    Find the common difference of an AP, whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.

Sol.    a₁ + a₂ + a₃ + a₄ = (a₅ + a₆ + a₇ + a₈)

⇒ 2[a₁ + a₂ + a₃ + a₄] = a₅ + a₆ + a₇ + a₈

⇒ 2[a₁ + a₂ + a₃ + a₄] + (a₁ + a₂ + a₃ + a₄) = [a₁ + a₂ + a₃ + a₄]+ (a₅ + a₆ + a₇ + a₈)

(adding both side a₁ + a₂ + a₃ + a₄)

⇒ 3(a₁ + a₂ + a₃ + a₄) = a₁ + …. + a₈ ⇒ 3S₄ = S₈

(Rightarrow 3left[ frac{4}{2}(2times 5+(4-1),,d right]=left[ frac{8}{2}(2times 5+(8-1),,d right])

⇒ 3[10 + 3d] = 2[10 + 7d]

⇒ 30 + 9d = 20 + 14d   ⇒   5d = 10 ⇒ d = 2

Example 19:    If the nth term of an AP is (2n + 1) then find the sum of its first three terms.

Sol.    ∵ a_n = 2n + 1

a₁ = 2(1) + 1 = 3

a₂ = 2(2) + 1 = 5

a₃ = 2(3) + 1 = 7

∴  a₁ + a₂ + a₃ = 3 + 5 + 7 = 15