**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 :

Number of terms | Terms to be taken |

3 | a – d, a, a + d |

4 | a – 3d, a – d, a + d, a + 3d |

5 | a – 2d, a – d, a, a + d, a + 2d |

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 2*r* terms in an A.P.

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

Number of terms | Terms to be taken |

3 | a, a + d, a + 2d |

4 | a, a + d, a + 2d, a + 3d |

5 | a, a + d, a + 2d, a + 3d, a + 4d |

### 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, A_{1}, A_{2}, A_{3}, ….. A_{n}, b are in A.P., then A_{1}, A_{2}, A_{3}, ….. 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 A_{1}, A_{2}, A_{3}, ….. A_{n} are n A.M.’s between a and b, then

### Properties of A.P.

- If a
_{1}, a_{2}, a_{3}, …… are in A.P. whose common difference is*d*, then for fixed non-zero number*k*∈*R*.- a
_{1}± k, a_{2 }± k, a_{3}± k, .….. will be in A.P., whose common difference will be*d*. - ka
_{1}, ka_{2}, ka_{3}, …… will be in A.P. with common difference =*kd*. - a
_{1}/k, a_{2}/k, a_{3}/k, ……will be in A.P. with common difference =*d*/*k*.

- a
- 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.*a_{1}+ a_{n}= a_{2}+ a_{n}_{−}_{1}= a3 + a_{n}_{−}_{2}= ….. - 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.
- 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.
- If the number of terms of an A.P. is odd then its middle term is A.M. of first and last term.
- If a
_{1}, a_{2},…… a_{n}and b_{1}, b_{2},…… b_{n}are the two A.P.’s. Then a_{1}± b_{1}, a_{2 }± b_{2},.….. a_{n}± b_{n}are also A.P.’s with common difference d_{1}≠ d_{2}, where d_{1}and d_{2}are the common difference of the given A.P.’s. - Three numbers
*a*,*b*,*c*are in A.P. iff 2b = a + c. - 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}. - 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**

^{th }

**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

∴ T

_{26}from last = T

_{(96-26+1) }from beginning

= T

_{71}from beginning

= a + 70d

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

**Method: II**

d = 15 – 7 = 8

∴ from last, a = 767 and d = –8

∴ T

_{26}= 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} = n^{2} + 1

**Solution.** ** (i)** We have,

a_{n} = 3n + 2

Putting n = 1, 2 and 3, we get

a_{1} = 3 × 1 + 2 = 3 + 2 = 5,

a_{2} = 3 × 2 + 2 = 6 + 2 = 8,

a_{3} = 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} = n^{2} + 1

Putting n = 1, 2, and 3 we get

a_{1} = 12 + 1 = 1 + 1 = 2

a_{2} = 22 + 1 = 4 + 1 = 5

a_{3} = 32 + 1 = 9 + 1 = 10

Thus, the first three terms of the sequence defined by a_{n} = n^{2} + 1 are 2, 5 and 10.

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

**Solution.** a_{n} = (–1)^{n-1} × 2^{n}

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

a_{1} = (–1)^{1-1} × 2^{1} = (–1)^{0} × 2 = 2

a_{2} = (–1)^{2-1} × 2^{2} = (–1)^{1} × 4 = – 4

a_{3} = (–1)^{3-1} × 2^{3} = (–1)^{2} × 8 × 8

a_{4} = (–1)^{4-1} × 2^{4} = (–1)^{3} × 16 = –16

a_{5} = (–1)^{5-1} × 2^{5} = (–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

a_{10} = 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,

a_{12} = a + (12 – 1) d = a + 11d = 9 + 11 × 4 = 53

a_{24} = a + (24 – 1) d = a + 23 d = 9 + 23 × 4 = 101

and, a_{n} = a + (n – 1) d = 9 + (n – 1) × 4 = 4n + 5

a_{12} = 53, a_{24} = 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

a_{1}, a_{2}, a_{3}, ….. an, ……

It is given that a_{10} = 52 and a_{16} = 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

∴ a_{32} = a + (32 – 1) d = 7 + 31 × 5 = 162

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

Hence a_{32} = 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, a_{2}, a_{3}, ……. a_{n}, …….

It is given that a_{7} = – 1 and a_{16} = 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 a_{1}, a_{2}, a_{3}, ….. , an, …. be the A.P. with its first term = a and common difference = d.

It is given that 5a_{5} = 8a_{8}

⇒ 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 ⇒ a_{13} = 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

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

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

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

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

∴ (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

4a^{2} + 20 d^{2} = 120

4 × 5^{2} + 20 d^{2} = 120

d^{2} = 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.** a_{1} + a_{2} + a_{3} + a_{4} = (a_{5} + a_{6} + a_{7} + a_{8})

⇒ 2[a_{1} + a_{2} + a_{3} + a_{4}] = a_{5} + a_{6} + a_{7} + a_{8}

⇒ 2[a_{1} + a_{2} + a_{3} + a_{4}] + (a_{1} + a_{2} + a_{3} + a_{4}) = [a_{1} + a_{2} + a_{3} + a_{4}]+ (a_{5} + a_{6} + a_{7} + a_{8})

(adding both side a_{1} + a_{2} + a_{3} + a_{4})

⇒ 3(a_{1} + a_{2} + a_{3} + a_{4}) = a_{1} + …. + a_{8} ⇒ 3S_{4} = S_{8}

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

⇒ 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

a_{1} = 2(1) + 1 = 3

a_{2} = 2(2) + 1 = 5

a_{3} = 2(3) + 1 = 7

∴ a_{1} + a_{2} + a_{3} = 3 + 5 + 7 = 15