5.2 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.

In (i), each term is 1 more than the term preceding it.

In (ii), each term is 30 less than the term preceding it.

In (iii), each term is obtained by adding 1 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) 0 to the term preceding it. In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the term preceding it.

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 ).

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 first term.

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

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

6, 9,12, 15, . . .

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

6, 3, 0, –3, . . .

Similarly, when

a = – 7, d = – 2, the AP is  – 7, – 9, – 11, – 13, . . .

a =  1.0, d = 0.1, the AP is  1.0, 1.1, 1.2, 1.3, . . .

a =  0, $$d = 1 \frac{1}{2}$$,  the AP is 0, 1 $$\frac{1}{2}$$, 3, 4 $$\frac{1}{2}$$, 6,..

a = 2, d = 0, the AP is  2, 2, 2, 2, . . .

Example 1 : For the $$AP:\frac{3}{2}, \frac{1}{2}, -\frac{1}{2}, -\frac{3}{2}$$,…..,write the first term a and the common difference d.

Solution : here, $$a=\frac{3}{2},  d=\frac{1}{2}-\frac{3}{2}=-1$$

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) 4, 10, 16, 22, . . .              (ii) 1, – 1, – 3, – 5, . . .

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

Solution : (i) We have $$a_2 – a_1 = 10 – 4 =   6$$

$$a_ 3 – a_ 2 = 16 – 10 =  6$$

$$a_ 4 – a _3 = 22 – 16  = 6$$

i.e., $$a_{K+1}-a_K$$  is the same every time.

So, the given list of numbers forms an AP with the common difference d = 6.

The next two terms are: 22 + 6 = 28 and 28 + 6 = 34.

(ii) $$a_2 – a1 =  – 1 – 1 = – 2$$

$$a_3 – a _2   =  – 3 – ( –1 ) = – 3 + 1 = – 2 $$

$$a_4 – a_ 3  = – 5 – ( –3 ) = – 5 + 3 = – 2$$

i.e., $$a_{K+1} – a_{k}$$  is the same every time.

So, the given list of numbers forms an AP with the common difference d = –2.

(iii) $$a_2 – a_1 = 2 – (– 2) = 2 + 2 = 4$$

$$a_3 – a_2  = – 2 – 2 = – 4 $$

As $$a_2 – a_1  ≠ a_3 – a_2$$ , the given list of numbers does not form an AP.

(iv) $$a_2 – a_1 = 1 – 1 = 0$$

$$a_3 – a_2 = 1 – 1 = 0$$

$$a_4 – a_3 = 2 – 1 = 1$$

Here, $$a_2 – a_1 = a_3 – a_2 ≠ a_4 – a_3$$.

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