# Solution

This was an arithmetic progression problem and much too simple. Even I could solve it without any effort. But this was just to get started with AP; we might solve more AP examples later. To solve this example, I had only to plug in the values in the formula.

If Praveen delayed the work by one day, he would have to pay a fine of Rs. 500. The fine would increase by Rs. 50 each succeeding day. So, the common difference, d = 50, and the first term of AP, a = 500. So, if Praveen were to delay the work indefinitely, he would have to pay 500 + 550 + 600 + 650 + 700 + 750 + ...

The formula to find the sum of AP is S = n/2 [2a + (n-1)d]. Now, "n" is the number of terms in the AP; in this case, Praveen has delayed the work for 30 days, so n= 30. And, of course, a = 500 and d = 50. So, let's plug in the values:

So, S = 30/2 * [2(500) + (30 - 1)*50]

So, S = 15 (1000 + 1450) = 36750.

Praveen had to pay Rs. 36,750 as the penalty. That was a huge loss!

Okay, so there was nothing much in this example. You had only to know the formula. But, in case you happen to forget the formula, you must know how to derive it. It is really smart how this formula is derived.

Suppose we have the following AP: 5, 8, 11, 14, 17. Here, the first term a = 5, the common difference d = 3, and the last term an = 17.

So, the first term, a = 5

The second term = 8 = (a + d)

The third term = 11 = (a + 2d)

The fourth term = 14 = (a + 3d)

The fifth and the last term = an = 17 = (a + 4d).

You will have noticed a pattern. If there are "n" terms, the last term an will be [a + (n - 1)d].

For our AP example, the sum is S = 5 + 8 + 11 + 14 + 17.

The sum will be the same even if we reverse the order of the terms.

So, S = 17 + 14 + 11 + 8 + 5. This is the same as: an + (an - d) + (an - 2d) + (an - 3d) + (an - 4d).

Well, that's all there to it.

Now, let's consider an AP of "n" terms: a, a+d, a+2d, a+3d, .............. a+(n-1)d. Please note the nth term is [a + (n - 1)d].

So, the sum, S = a + (a + d) + (a + 2d) + (a + 3d) + ....................... + [a + (n - 1)d]. Let me label this as Equation I.

We know that the sum will be the same even if the terms are reversed.

So, S = an + (an - d) + (an - 2d) + (an - 3d) + ........... + [(an - (n - 1)d)]. Please note, here the last term will be [(an - (n - 1)d)]. Let me label this as Equation II.

If I add Equation I and Equation II, I get: S + S = (a + an) + (a + an) + (a + an) + ......... + (a + an). All the "d" terms get cancelled out.

Therefore, 2S = n * (a + an). This is because (a + an) is added "n" times.

So, S = n/2 * (a + an).

But, we know that the nth term of an AP is an = [a + (n - 1)d]. Let's substitute this in the equation.

We get,S = n/2 [a + a + (n - 1)d]. Therefore, S = n/2 [2a + (n - 1)d].

What a great way to find the sum of an AP! There may be hundreds of terms, but we can get the sum by applying the formula.

Suppose I want to add up all the natural numbers from 1 to n. This is an AP where a = 1, d = 1, and there are "n" terms. If I plug in the values, I get S = n/2 * (n + 1). If n = 2000, I get S = 2000/2 * (2000 + 1) = 2001000. I have added up all the natural numbers from 1 to 2000 in just one step!

• Union Public Service Commission - www.upsc.gov.in
• IIT-Kharagpur - www.iitkgp.ac.in
• Indian Statistical Institute - www.isical.ac.in
• Indian Institute of Technology Madras - www.iitm.ac.in
• Indian Institute of Management, Ahmedabad - www.iimahd.ernet.in
• Indian Institute of Mass Commission - www.iimc.nic.in
• IIT Bombay - www.iitb.ac.in