In this post we will study the basic concept of Arithmetic Progression which is relevant for Grade 11 Mathematics syllabus. Apart from the concept we will also study the formulas and solved problems which are important part of CBSE/NCERT syllabus.

**Arithmetic Progression Concept**

If the elements are arranged in following sequence:

a, a + d, a + 2d, a + d, a + 4d, a + 5d, . . . . . . . a + (n-1)d

then the sequence is in the form of Arithmetic Progression (AP).

Here,

a is the first term

d is the common difference

In Arithmetic Progression

(a) The difference of consecutive numbers is always same (i.e. d)

(b) this common difference can help us to find other unknown numbers in the series

**Example 01****2, 4, 6, 8, 10, 12, . . . . . .**

This sequence is in the form of arithmetic progression because:

(a) the common difference between consecutive term is same (d = 2)

(b) The sequence can also be represented in AP form**a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, . . . . . . . a + (n-1)d****2, 2+2, 2 +(2.2), 2+ (3.2), 2+ (4.2), 2+ (5.2) . . . . . . . 2 + (n-1)2**

**Example 02****7, 14, 21, 28, 35, 42, . . . . . . . . . . . . . **

The sequence is in the form of Arithmetic Progression because the difference between consecutive numbers is same

**Example 03****4, 0, -4, -8, -12, -16, -20, -24, -28, . . . . . . . . . . . .**

The sequence is Arithmetic Progression because common difference between elements is same

**Formula for Arithmetic Progression**

1. **nth term of Arithmetic Progression****Tn = a + (n – 1) d**

Tn = Value of nth term of the AP

a = First Term of AP

d = Common difference of AP

2. **Sum to nth term of AP** Sn\ =\ \frac{n}{2} \ [ 2a\ +\ ( n-1) d]

Sn = Sum to nth terms of AP

a = First term of AP

d = common difference

3. **Sum to nth term when the last term is given**

Sn\ =\ \frac{n}{2} \ [ a+l]

Sn = Sum to nth terms of AP

a = First term of AP

l = last term of AP

**4. Arithmetic Mean**

Given two numbers a & b, we can calculate the arithmetic mean of numbers using the formula:**A = (a + b)/2**

The calculated Arithmetic mean (A) is such that the sequence **a, A, b** is in Arithmetic Progression

**Example**

Let there are two number 2 and 10.

Arithmetic Mean = (2 + 10)/2

Arithmetic Mean = 12/2 = 6

Now the sequence 2, 6, 10 forms Arithmetic Progression as they have same common difference.**6 – 2 = 4****10 – 6 = 4**

You can see that common difference is 4

**Arithmetic Progression Questions**

**(01) The income of a person is $3,00,000 in the first year and he receives an increase of $10,000 to his income per year for the next 19 years. Find the total amount, he received in 20 years**

Solution

Income First year = 3,00,000

Income Second Year = 3,00,000 +10,000 =3,10,000

Income Third Year = 3,10,000 +10,000 = 3,20,000

Income Fourth Year = 3,20,000 + 10,000 = 3,3000

You can see that the sequence is forming Arithmetic Progression with common difference of 10,000

The sequence is like**300000**, **310000**, **320000**, **330000**, **340000**, . . . . . . . . . . . . . . . . . . . .

Let us find the amount the person received in 20th year

The 20th term in the above series is:

Tn = a + (n – 1)d

T20 = 300000 + (20-1) * 10,000

T20 = 300000 + 19 * 10,000

T20 = 300000 + 190000

T20 = $4,90,000

The person will receive $4,90,000 in the 20th year

So the sequence becomes**300000**, **310000**, **320000**, **330000**, **340000**, . . . . . . . . . . . . . . . . . . .**490000**

Sum of all the amount is calculated using formula:

Sn = n/2 (a + l)

Here,

n= 20

a = 3,00,000

l = 4,90,000**Putting the values**

S20 = 20/2 * (3,00,000 + 4,90,000)

S20 = 10 * (7,90,000)

S20 = $ 79,00,000

**Hence the total sum till 20th years is $79,00,000**

**(02) The 6th and 20th terms of an AP are 8 and −20 respectively. Find the 30th term**

Solution

Formula for nth term of AP is:

Tn = a + ( n – 1)d

Using the above formula, 6th Term of AP can be represented as ==> a + 5d

and 20th term can be represented as ==> a+19d

As given in the question:**a + 5d = 8** —– eq (1)**a + 19d = -20** —— eq (2)

Equating eq(2) – eq(1), we get

a + 19d – a – 5d= – 20 – 8

14 d = -28**d = – 2**

Putting the value of d in eq(1), we get

a + 5 (-2) = 8

a – 10 = 8**a = 18**

Now the 30th term will be

T30 = a + 29 (d)

T30 = 18 + 29 (-2)

T30 = 18 – 58**T30 = – 40**

Hence the 30th term is **-40**

**(03) There is an AP 11, 13, 15…. Which term of this AP is 65?**

**Solution**

The following sequence is in Arithmetic Progression

11, 13, 15, 17, 19 . . . . .65 . . . . .

Here

First Term (a) = 11

Common Difference (d) = 13 – 11 = 2

We have to find the term whose value is 65

Using formula

Tn = a + (n – 1)d

65 = 11 + (n – 1)*2

54/2 = n -1

27 = n – 1

n = 28**Hence the 28th term of above AP is 65**