Definition :- An Arithmetic Progression is a sequence in which the difference between a term and it’s preceding term is a constant. The constant is the common difference (c.d) and is denoted by ‘d’. (The Arithmetic Progression is abbreviated as A.P).

**The general form and term of an A.P :**

T1, T2, T3 ………….. Tn .......... be an A.P

T2 –T1 = T3 –T2 = T4 –T3 = ………….. = Tn – Tn-1 = d

Let T1 = a = a + (1 – 1) d

∴ T2 = T1 + d = a + d = a + (2 – 1)d

T3 = T2 + d = a + 2d = a + (3 – 1)d

T4 = T3 + d = a + 3d = a + (4 – 1)d

………………………………….

Tn = Tn-1 + d = a + (n – 1)d

∴**a, a + d, a + 2d, a + 3d, ...................., a + (n** – **1)d** is the general form of an

arithmetic progression with ‘a’ as the first term and having common difference ‘d’. This is the Standard form of an A.P. The last term a + (n – 1) d is denoted by Tn or *l*

**Note :** 1. In an A.P succeeding term of a given term is obtained by adding ‘d’to it. [Tn+1 = Tn+ d].

2. In an A.P preceding term of a given term is obtained by subtracting ‘d’ from it [Tn-1 = Tn – d].

3. Tn = a + (n – 1)d is the general term of an A.P

Find the nth and the 20th terms of the A.P. 3, 7, 11, 15, ...............

**Solution :** a = 3, d = T2 – T1 = 7 – 3 = 4, Tn = ?, T20 = ?

Tn = a + (n – 1)d.

Tn = 3 + (n – 1)4

= 3+4n–4

**Tn = 4n** – **1**

T20 = 4(20) – 1

T20 = 79

(OR) T20 = 30 + (20 – 1)4

= 80 – 1 = 3 + 19 x 4

T20 = 3+76=79

How many terms are there in the A.P. 4, –1, –6, ............., (–106)

**Solution :** Given : a = 4, d = –5, Tn = –106, n = ?

Tn = a + (n – 1)d.

∴ a + (n – 1)d = Tn

4 + (n – 1) (–5) = –106

4 – 5n + 5 = –106

–5n + 9 = –106

∴ –5n = –106 – 9

∴ –5n = –115

n = 23

There are 23 terms in the given A.P.

Q . The fourth and eighth terms of an A.P. are in the ratio 1 : 2 and tenth term is 30, Find the Common difference and the first term.[ d = 3; a = 3]

Q. The angles of a triangle are in A.P. The smallest angle is 300. Show that the triangle is a right angled triangle.[ In triangle ABC, let us say ∠A = 30 ^{0} then <∠B = 30^{0}+d and <∠C = 30^{0} +2d ]

Q. Find the three numbers of an A.P. whose sum is 12 and their product is 48.

[2, 4 and 6 (or) 6, 4 and 2]

Q. Find the sum of all positive multiples of 3 less than 50 [408]

**Sum of First n Terms of an AP**

Let us denote the first term of an AP by a_{1}, second term by a_{2} , . . ., nth term by a_{n} and the common difference by d. Then the AP becomes a_{1}, a_{2}, a_{3}. . . a + [n-1]d

Let S denote the sum of the first n terms of the AP. We have

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

Rewriting the terms in reverse order, we have

S = [*a* + (*n* – 1) *d* ] + [*a* + (*n* – 2) *d* ] + . . . + (*a* + *d* ) + *a ----------------*(ii)

On adding (i) and (ii), term-wise. we get

2S = [2*a* + (*n* – 1) *d* ] + [2*a* + (*n* – 1) *d* ] + . . . + [2*a* + (*n* – 1) *d* ] ----- n times

S = [2a + (n – 1) d ]

OR, S = [a + a + (n – 1) d ] = [a + a_{n} ]

OR, if there are only n terms in an AP, then an = l, the last term.

S = (a + l )

This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.

**Find the sum of all natural numbers between 91 and 170 which are divisible by 5.**

**Solution :** The numbers between 91 and 170 which are divisible by 5 are 95, 100, 105, ...... 165

a = 95 ; d = 5

First find the number of terms in the A.P.

Tn = 165

∴ a + (n – 1)d = 165

95 + (n – 1)5 = 165

∴ 95 + 5n – 5 = 165

∴ 5n + 90 = 165

5n = 165 – 90

n = 15

Sn = n/2 (a + *l*) =15/2 (95 + 165) =1950

**The sum of n terms of an arithmetic series is Sn = 2n ^{2} + 6n. Find the first term and the common difference.**

**Solution :** Given : Sn = 2n2 + 6n and S1 = T1

S1 = 2(1)^{2} + 6(1) = 2+6 = 8

T1 = 8

∴ a = 8

To find 'd'

S2 = 2(2)^{2} + 6(2) = 2(4) + 12 = 8 + 12

∴ S2 = 20

S2 = T1 + T2 = 20

20 = 8 + T2

T2 = 12

d = T2 – T1 = 12 – 8

Q. How many terms of the series 10 + 8 + 6 + ............ should be added to get the sum –126. [n = 18 ]

Q. Sanganbasava rides a bicycle from his home to the ashram. He covers 125 meters at the end of first minute, 135 meters at the end of second minute and so on. If he reaches the Ashram at the end of 10 minutes. Find the distance between hishome and ashram. [1.7 Kms]

Q. Find the A.P in which a)Tn = 2n + 1 b) Tn = 4 – 5n c) Sn = 5n^{2} + 3n

[I. a) 3, 5, 7, ............. b) –1, –6, –11, .............. c) 8, 18, 28 ............]

Q. Find the sum of the Arithmetic series which contains 25 terms and whose middle term is 20. [500]

Q. The angles of a quadrilateral are in A.P. If the smallest angle is 15^{0}, find the angles of the quadrilateral.[ 65^{0}, 115 ^{0} and 165^{0}]

Q. Veershree climbed 23 steps of Golgumbaz in the first minute. After that she climbed two steps less than what she had climbed in the previous minute. If she reached the whispering gallery of Golgumbaz after 7 minutes how many steps did she climb to reach the whispering gallery? [119]

**Arithmetic progressions**