I am sure most of us have used a taxi to reach a destination. So, you know that a taxi charges a base fare for the first few kilometres. After which, it charges a certain amount per kilometre. For example, say the base fare of a taxi is 20 Rs for the first 2 km. After which, taxi charges 10 Rs per km. So charges for 3, 4, 5…km will be 30, 40, 50….Observe that these taxi fares form a pattern. In which, we get the next fare by adding a fixed number to the previous fare. Such a sequence of numbers is said to be in Arithmetic Progression. Let us learn more about arithmetic progressions.

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## What is a Sequence?

A sequence or progressions is a list of numbers in a special order. It is a string of numbers following a particular pattern, and all the elements of a sequence are called its terms. There are various types of sequences which are universally accepted, but the one which we are going to study right now is the arithmetic progression.

## Arithmetic Progression (AP)

An arithmetic progression or arithmetic sequence is a sequence in which the difference between any two consecutive terms is constant. The difference between the consecutive terms is known as the common difference and is denoted by **d**. Let us understand this with one example.

Let’s check whether the given sequence is A.P: 1, 3, 5, 7, 9, 11. To check if the given sequence is in A.P or not, we must first prove that the difference between the consecutive terms is constant. So, d = a_{2}– a_{1 }should be equal to a_{3}– a_{2 }and so on… Here,

d = 3 – 1 = 2 equal to 5 – 3 = 2

Here we can see that the difference is common to both the terms. So, we can say that the given sequences are in arithmetic progression.

**Browse more Topics under Sequences And Series**

**General Form of an Arithmetic Progression**

Say the terms a_{1 } a_{2 }a_{3}……a_{n }are in AP. If the first term is ‘a’ and its common difference is ‘d’. Then, the terms can also be expressed as follows

1^{st} term a_{1} = a

2^{nd} term a_{2} = _{ }a_{ }+ d

3^{rd} term a_{3 = }a + 2d

Therefore, we can also represent arithmetic progressions as

**a, a + d, a + 2d, ……**

This form of representation is called the general form of an AP.

**n**^{th} term of an Arithmetic Progression

^{th}term of an Arithmetic Progression

Say the terms a_{1 } a_{2 }a_{3}……a_{n }are in AP. If the first term is ‘a’ and its common difference is ‘d’. Then, the terms can also be expressed as follows,

2^{nd} term a_{2} = a_{1}+ d = _{ }a_{ }+ d = a + (**2-1**) d

3^{rd} term a_{3 = }a_{2 }+ d = (a + d) + d = a + 2d = a + (**3-1**)d

likewise, n^{th }term a_{n} = a + (n-1) d

Therefore, we can find the nth term of an AP by using the formula,

**a _{n }= (a + (n – 1) d)**

a_{n }is called the general term of an AP

**Sum of n terms in an Arithmetic Progression**

The sum of first n terms in arithmetic progressions can be calculated using the formula given below.

**S = [(n/2) * (2a + (n – 1) d)] **

Here S is the sum, n is the number of terms in AP, a is the first term and d is the common difference.

When we know the first term, a and the last term, l, of AP. Then, the sum of n terms is

**S** = **[(n/2) * (a+l)] **

## Properties of Arithmetic Progressions

- If the same number is added or subtracted from each term of an A.P, then the resulting terms in the sequence are also in A.P with the same common difference.
- If each term in an A.P is divided or multiply with the same non-zero number, then the resulting sequence is also in an A.P
- Three number x, y and z are in an A.P if 2y = x + z
- A sequence is an A.P if its n
^{th }term is a linear expression. - If we select terms in the regular interval from an A.P, these selected terms will also be in AP

## Solved Problems for You

**Multiple Choice Questions**

Question: If l = 20, d = -1 and n = 17, then the first term is?

- 30
- 32
- 34
- 36

Solution: The correct option is D

Given: The last term of AP refers to as l. Given that l = 20, d = -1, n = 17. We need to find *a. *

By using l = a + (n – 1)d, we can find a.

20 = a + (17 – 1) -1

20 = a – 16

a = 36

Question: For an essay writing competition, a total of four entries received a cash prize. The first prize was Rs 2000. Each prize is Rs 500 less than the first prize. What was the total sum of cash given out to students for the essay writing competition?

- 4500
- 5000
- 5500
- 4000

Solution: Correct option is B.

As per given data, n = 4, a = 2000 and d = – 500. By using the formula S = [(n/2) * (2a + (n – 1) d)] we can calculate the total cash given out to winners. On subtitution we get,

S = [(4/2) * (2(2000) + (4 – 1) (-500)] = [(2) * (4000 + (3) (-500)] = [(2) * (4000 + (-1500)] = [(2) * (2500)] = 5000

Hence, a total sum of Rs 5000 was handed out in cash to winners.

**Answer The Following**

** Question.** Write first five terms of AP, when a = 23 and d = -5

**Answer.** The general form of AP is a, a+d, a+2d …

When a = 23 and d = -5, the first five terms of AP are 23, 23+(-5), 23 + 2(-5), 23 +3(-5), 23 + 4(-5)

Hence, the first five terms of AP are 23, 18, 13, 8 and 3

**Question.** Find the 50^{th} odd number.

**Answer.** Odd numbers are in AP. In which, a = 1 and d = 2. So, using the formula a_{n }= (a + (n – 1) d) we can find the 50^{th} odd number.

By substituting in the formula we get

a_{50} = 1 + (50-1)2

= 1 + (49)2 = 1 + 98 = 99

Thus, the 50^{th} odd number is 99.

**Question.** Find the missing term in the following AP. 5, x, 13

**Answer.** Three numbers x, y, z are in AP if 2y = x + z. Given x = 5, z = 13 therefore y = (x+z)/2.

By substituting values we get

y = (5+13)/2 = 18/2 = 9

Therefore, the missing term is 9.

**Question.** Find the sum of first 50 natural numbers.

**Answer.** By using formula, S = [(n/2) * (a+l)] we can find the sum. On substitution, we get

= [(50/2) * (1+50)] = [(25) * (51)] = 1275

Therefore, sum of first 50 natural numbers is 1275.