5 - ARITHMETIC PROGRESSIONS

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 of ₹ 8000 after 3, 6, 9 and 12 years will be, respectively: 10000, 12500, 15625, 19531.25.
(iv) The number of unit squares in squares with side 1, 2, 3, ... units (see Fig. 5.2) are, respectively 12, 22, 32, ... .
(v) Shakila puts ₹ 100 into her daughter’s money box when she was one year old and increased the amount by ₹ 50 every year. The amounts of money (in ₹) in the box on the 1st, 2nd, 3rd, 4th, ... birthday were 100, 150, 200, 250, ..., respectively.
(vi) A pair of rabbits are too young to produce in their first month. In the second, and every subsequent month, they produce a new pair. Each new pair of rabbits produce a new pair in their second month and in every subsequent month (see Fig. 5.3). Assuming no rabbit dies, the number of pairs of rabbits at the start of the 1st, 2nd, 3rd, ..., 6th month, respectively are: 1, 1, 2, 3, 5, 8.

In the examples above, we observe some patterns. In some, we find that the succeeding terms are obtained by adding a fixed number, in others by multiplying with a fixed number, in another we find that they are squares of consecutive numbers, and so on.

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.

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 a , second term by a , ..., nth term by a and the common difference by d. Then the AP becomes a , a , a , ..., a . So, a – a = a – a = ... = a – a = 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.

Let a , a , a , . . . be an AP whose first term a is a and the common difference is d. Then, the second term a = a + d = a + (2 – 1) d, the third term a = a + d = (a + d) + d = a + 2d = a + (3 – 1) d, the fourth term a = a + d = (a + 2d) + d = a + 3d = a + (4 – 1) d, ... So, the nth term a of the AP with first term a and common difference d is given by a = 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]. Rewriting the terms in reverse order, we have: S = [a + (n – 1) d] + [a + (n – 2) d] + ... + (a + d) + a. Adding these two equations gives: 2S = n [2a + (n – 1) d]. Therefore, the sum of the first n terms of an AP is given by: S = (n / 2) * (2a + (n - 1)d).