ARITHMETIC PROGRESSIONS

CHAPTER -8

ARITHMETIC PROGRESSIONS

Introduction

You must have observed that in nature, many things follow a certain pattern, such as the petals of a sunflower, the holes of a honeycomb, the grains on a maize cob, the spirals on a pineapple and on a pine cone etc.
We now look for some patterns which occur in our day-to-day life. Some such examples are:
(i) Reena applied for a job and got selected. She has been offered a job with a starting monthly salary of   8000, with an annual increment of 500 in her salary. Her salary (in  ) for the 1st, 2nd, 3rd, . . . years will be, respectively
8000, 8500, 9000, . . . .
(ii) The lengths of the rungs of a ladder decrease uniformly by 2 cm from bottom to top (see  Fig. 5.1). The bottom rung is 45 cm in length. The lengths (in cm) of the 1st, 2nd, 3rd, . . ., 8th rung from the bottom to the top are, respectively
45, 43, 41, 39, 37, 35, 33, 31
(iii) In a savings scheme, the amount becomes   times of itself after every 3 years.       
The maturity amount (in) of an investment 8000 after 3, 6, 9 and 12 years will be, respectively:
10000, 12500, 15625, 19531.25

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 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.
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).
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.
Let us denote the first term of an AP by a₁, second term by a₂, …..,nth term by a_n and the common difference by d. then the AP becomes a₁, a₂, a₃, ….., a_n.
So, a₂ – a₁ = a₃ – a₂ = … = a_n – a_{n – 1}= d.
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 degree celsius) recorded for a week in the month of January in a city, arranged in ascending order are
– 3.1, – 3.0, – 2.9, – 2.8, – 2.7, – 2.6, – 2.5
(c) The balance money (in) after paying 5 % of the total loan of   1000 every month is 950, 900, 850, 800, . . ., 50.
(d) The cash prizes (in) given by a school to the toppers of Classes I to XII are, respectively, 200, 250, 300, 350, . . ., 750.
(e) The total savings (in) after every month for 10 months when   50 are saved each month are 50, 100, 150, 200, 250, 300, 350, 400, 450, 500.
It is left as an exercise for you to explain why each of the lists above is an AP.
You can see that
a, a + d, a + 2d, a + 3d, . . .
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.

nth Term of an AP
Let us consider the situation again, given in Section 5.1 in which Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of   8000, with an annual increment of    500. What would be her monthly salary for the fifth year?
Let a₁, a₂, a₃,…be an AP whose first term a₁ is a and the common difference is d.
Then,
The second term a₂ = a + d = a + (2 – 1) d
The third term     a₃ = a₂ + d = (a + d) = a + 2d = a + (3 – 1) d
The fourth term   a₄ = a₃ + d = (a + 2d) + d = a + 3d = a + (4 – 1) d…………..
Looking at the pattern, we can say that the nth term a_n = a + (n – 1) d.
So, the nth term a_n of the AP with first term a and common difference d is given by
a_n = a + (n – 1) d.
a_n is also called the general term of the AP. If there are m terms in the AP, then a_m represents the last term which is sometimes also denoted by l.

Sum of First n Terms of an AP

Let us consider the situation again given in Section 5.1 in which Shakila put   100 into her daughter’s money box when she was one year old, 150 on her second birthday, 200 on her third birthday and will continue in the same way. How much money will be collected in the money box by the time her daughter is 21 years old?

Here, the amount of money (in  ) put in the money box on her first, second, third, fourth . . . birthday were respectively 100, 150, 200, 250, . . . till her 21st birthday. To find the total amount in the money box on her 21st birthday, we will have to write each of the 21 numbers in the list above and then add them up. Don’t you think it would be a tedious and time consuming process? Can we make the process shorter? This would be possible if we can find a method for getting this sum. Let us see.
We  consider  the  problem  given  to  Gauss  (about  whom  you  read  in Chapter 1), to solve when he was just 10 years old. He was asked to find the sum of the positive integers from 1 to 100. He immediately replied that the sum is 5050. Can you guess how did he do? He wrote:
S = 1 + 2 + 3 + . . . + 99 + 100
And then, reversed the numbers to write
S = 100 + 99 + . . . + 3 + 2 + 1

Adding these two, he got
2S = (100 + 1) + (99 + 2) + . . . + (3 + 98) + (2 + 99) + (1 + 100)
= 101 + 101 + . . . + 101 + 101  (100 times)
So,      i.e., the sum = 5050.
We will now use the same technique to find the sum of the first n terms of an AP:
a, a + d, a + 2d, . . .
The nth term of this AP is 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]                        (1)
Rewriting the terms in reverse order, we have
= [a + (n – 1) d] + [a + (n – 2) d] + . . . + (a + d) + a                  (2)
On adding (1) and (2), term-wise. We get

or,     2S = n [2a + (n – 1) d]           (Since, there are n terms)

or,   

So, the sum of the first n terms of an AP is given by                   

We can also write this as   

i.e.,                                                       (3)

Now, if there are only n terms in an AP, then a_n = l, the last term. From (3), we see that
                                            (4)