Arithmetic Progressions

Consider the following lists of numbers:

(i)

1, 2, 3 , 4 , …

(ii)

100, 70, 40 , 10 , …

(iii)

-3, -2, -1 , 0 , …

(iv)

3, 3, 3 , 3 , …

(v)

-1.0, -1.5, -2.0 , -2.5 , …

Each of the numbers in the list is called a term.

Given a term, can you write the next term in each of the lists above? If so, how will you write it? Perhaps by following a pattern or rule. Let us observe and write the rule.

In (i), each term is more than the term preceding it.

In (ii), each term is less than the term preceding it.

In (iii), each term is obtained by adding to the term preceding it.

In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding (or subtracting) to the term preceding it.

In (v), each term is obtained by subtracting from the term preceding it.

In all the lists above, we see that successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form an Arithmetic Progression (AP).

An arithmetic sequence has a constant difference d between consecutive terms.

The same number is added or subtracted to every term, to produce the next one.

So, an arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the term.

This fixed number is called the common difference of the AP. Remember that it can be , or .

Let us denote the first term of an AP by a1 , second term by a2, . . ., nth term by an and the common difference by d. Then the AP becomes:

a1, a2, a3 , . . ., an.

So, a₂ – a₁ = a₃ – a₂ = . . . = a_n – a_{n – 1} =

Some more examples of AP are:

(a) The heights ( in cm ) of some students of a school standing in a queue in the morning assembly are 147 , 148, 149, . . ., 157.

(b) The minimum temperatures ( in °C ) recorded for a week in the month of January in a city, arranged in ascending order are:

– 3.1 °C, – 3.0 °C, – 2.9 °C, – 2.8 °C, – 2.7 °C, – 2.6 °C, – 2.5 °C

(c) The balance money (in Rs.) after paying 5 % of the total loan of Rs. 1000 every month is 950, 900, 850, 800, . . ., 50.

(d) The cash prizes (in Rs.) given by a school to the toppers of Classes I to XII are, respectively, 200, 250, 300, 350, . . ., 750.

(e) The total savings (in Rs.) after every month for 10 months when Rs. 50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.

We can see that:

a, a + , a + 2d, a + , . . .

represents an arithmetic progression where a is the first term and d the common difference. This is called the general form of an AP.

Note that in examples (a) to (e) above, there are only a finite number of terms. Such an AP is called a finite AP. Also note that each of these Arithmetic Progressions (APs) has a last term. The APs in examples (i) to (v) in this section, are not finite APs and so they are called infinite Arithmetic Progressions. Such APs do not have a last term.

Now, to know about an AP, what is the minimum information that we need? Is it enough to know the first term? Or, is it enough to know only the common difference? We find that both are needed – the first term a and the common difference d.

For instance if the first term a is 6 and the common difference d is 3, then the AP is 6, 9, , 15, . . .

and if a is 6 and d is –3, then the AP is 6, , 0, , . . .

Similarly, when:

a = – 7, d = – 2, the AP is:

-7, -2, -11 , , … Pattern: “Add -2 to the previous number to get the next one.”

a = 1.0, d = 0.1, the AP is:

1.0, 1.1 0.1, , 1.3 , … Pattern: “Add 0.1 to the previous number to get the next one.”

a = 0, d = 112 , the AP is:

0, 1 12 1 12, , 4 12, ,… Pattern: “Add 1 (1/2) to the previous number to get the next one.”

a = 2, d = 0, the AP:

2, 0, 2 , , … Pattern: “Add 0 to the previous number to get the next one.”

So, if we know what a and d are, we can list the AP. What about the other way round? That is, if we are given a list of numbers can we say that it is an AP and then find a and d?

Since a is the first term, it can easily be written. We know that in an AP, every succeeding term is obtained by adding d to the preceding term. So, d found by subtracting any term from its succeeding term, i.e., the term which immediately follows it should be same for an AP.

For example, for the list of numbers :

6, 9, 12 , 15 , …

We have a₂ – a₁ = 9 – 6 = ,

a₃ – a₂ = – = ,

a₄ – a₃ = – = 3

In general, for an AP a1, a2,.....an we have

d = ak1 - ak

where ak1 and ak are the ( k + 1)th and the kth terms respectively.

To obtain d in a given AP, we need not find all of a2 - a1,a3- a2, a4- a3.... It is enough to find only one of them.

Can you find their patterns and calculate the next two terms?

3, 6 +3, 9 , 12 , 15 , , … Pattern: “Add 3 to the previous number to get the next one.”

4, 10 , 16 , 22 , 28 , , , … Pattern: “Add 6 to the previous number to get the next one.”

Let's try out some problems.

Example 1 : For the AP : 32, 12, – 12, – 32, . . ., write the first term a and the common difference d.

Solution : Here, a (first term )= ,

d (common difference) = 12 - 32 = .

Remember that we can find d using any two consecutive terms, once we know that the numbers are in AP.

Example 2 : Which of the following list of numbers form an AP? If they form an AP, write the next two terms :

(i)

(i) 4, 10, 16, 22, . . .

Solution :

We have We a₂ – a₁ = – =

a₃ – a₂ = – =

a₄ – a₃ = – =

Since, the difference between consecutive terms is equal, the list forms an AP.

Here, a (first term ) = with d (common difference) = .

The next two terms are: and .

(ii)

(ii) 1, – 1, – 3, – 5, . . .

Solution :

We have a₂ – a₁ = – =

a₃ – a₂ = – =

a₄ – a₃ = – =

Since, the difference between consecutive terms is equal, the list forms an AP.

Here, a (first term ) = with d (common difference) = .

The next two terms are: and .

(iii)

(iii) – 2, 2, – 2, 2, – 2, . . .

Solution :

We have We a₂ – a₁ = – =

a₃ – a₂ = – =

Since, a₂ – a₁ ≠ a₃ – a₂, the given list of numbers does not form an AP.

(iv)

(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

Solution:

a₂ – a₁ = – =

a₃ – a₂ = – =

a₄ – a₃ = – =

Here, a₂ – a₁ = a₃ – a₂ ≠ a₄ – a₃

So, the given list of numbers does not form an AP.

So, to summarise:

An arithmetic sequence has first term a and common difference d between consecutive terms.

Recursive formula: xn=xn−1+d

Explicit formula: xn=a+d×n−1

Using the arrows, vary the values of 'a' and 'd', to study the behaviour of the graph.

Arithmetic Sequence

a = ${a}, d = ${d}

${arithmetic(a,d,0)}, ${arithmetic(a,d,1)}, ${arithmetic(a,d,2)}, ${arithmetic(a,d,3)}, ${arithmetic(a,d,4)}, ${arithmetic(a,d,5)}, …

Notice how all arithmetic sequences look very similar: if the difference is positive, they steadily , and if the difference is negative, they steadily .

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