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Basic Principle of AP:

Progressions and series constitute an important part of the mathematics syllabus of IIT JEE. Arithmetic progression is the easiest of the three progressions. With a bit of hard work it becomes quite simple to have a good command over the concepts of A.P. 

What exactly do we mean by arithmetic progression?

Arithmetic progression is basically a sequence in which the difference between any two terms of the sequence is constant. For example the sequence of the form a, a+d, a+2d, …. is called an arithmetic progression.

Here ‘a’ is called the first term while ‘d’ is the common difference. The nth term of the A.P. is given by

an = a+(n-1)d.

The sum of first n terms is given by

Sn = n/2 [2a + (n-1) d]

View the video for more on arithmetic progression

Now, we move on to the basic principles of A.P.:

1. Addition/Subtraction of constant number to each term of an A.P. also results an A.P.

e.g. suppose a1, a2, a3, …, an are in A.P.

then a1 + k, a2 + k, a3 + k ……, an + k and

a1 –k, a2 – k, a3 – k, ……, an–k will also be in A.P.

where k ∈ R.

2. Multiplication/Division by a constant number to each term of an A.P. also results an A.P.

e.g. suppose a1, a2, a3, …, an are in A.P.

then ka1, ka2, ka3 ……, kan and

a1/k, a2/k, a3/k will also be in A.P.

where k ∈ R and k ≠ 0.

3. Addition/Subtraction of two A.P.’s also results an A.P.

e.g. suppose a1, a2, a3, …, an and b1, b2, b3, …, bn, are in A.P.

then a1+b1, a2+b2, a3+b3, ……, an+bn.

a1–b1, a2–b2, a3–b3, ……, an–bn will also be in A.P.

4. Reversing the order of an A.P. also results an A.P.

e.g. Suppose a1, a2, a3, ……, an are in A.P.

then an, an–1, ……, a3, a2, a1 will also be in A.P.

Some key points to remember: 

1. If nth term of an series is tn = An + B, then

 tn - tn–1 = An + B – A(n–1)–B

 = A = constant

 i.e. the series is in A.P.

 2. If tn = An2 + Bn + c, then

 tn – tn–1 = A(n2) + Bn + C – (A(n–1)2 + B(n–1) + c)

 = A (2n + 1) + B = ∪(n) (say)

 Then Un – Un–1 = 2A

  which is a constant.

This shows that if the difference of the terms of a series is constant, then the nth term is quadratic in n.

3. Similarly, if the difference the terms of a series is constant then the nth term of the series is a cubic in n i.e.

tn = An3 + Bn2 + Cn + D and so on.

4. If a fixed number is added or subtracted from each term of a given A.P., then the new resulting sequence is also an A.P. and with the same common difference. But in case the A.P. is multiplied by a fixed number k or divided by a non-zero fixed number k then the resulting sequence is an A.P. with the common difference multiplied by k.

5. If a1, a2, a3, …… and b1, b2, b3, …… are two A.P.’s with common differences d and d’ respectively then a1 + b1, a2 + b2, + a3 + b3, … is also an A.P. with common difference d + d’.

6. If a1, a2, a3, ……, an are in A.P., then a1 + an = a2 + an–1 = a3 + an–2 = …… and so on.

Another form of question that is generally asked in the exam is on finding three, four or five terms of an A.P. satisfying certain condition. It becomes simple to solve such questions if the terms re assumed in a particular manner as described below:

  • Three numbers are in A.P.:

Assume the numbers as α–ß, α, α+ß (here first term is α–ß and c.d. is ß).

  • Four numbers are in A.P.:

Assume the numbers as α–3ß, α–ß, α + ß, α+3ß (here first term is α-3ß and c.d. is 2ß.

  • Five convenient numbers in A.P.

The numbers are α–2 ß, α–ß, α, α + ß, α + 2 ß.

  • In general, we take a – rd, a – (r – 1)d, …., a – d, a, a + rd in case we have to take (2r + 1) terms in an A.P. 

Basically sum of the above sets of variables eliminate one variable which make easy to solve the problems.

Illustration: In a triangle, the lengths of the two larger sides are 10 and 9 respectively. If the angles are in A.P, then find the length of the third side.

Solution: Let us assume the angles of the triangle as a, a-d, a+d. (since there are three angles in a triangle so the three numbers in A.P. should be assumed in this way)

Since the angles of the triangle are given to be in A.P, so we have

a + (a-d) + (a+d) = 180°

Hence a = 60°.

Now by using the cosine law in ?ACB, cos 60° = [(10)2 + 92 - x2]/2.10.9

This implies 1/2 = (181-x2)/ 180

So x2 = 91 gives x = √91.

Illustration: If the sum of the first 2n terms of the A.P. series 2, 5, 8, …. is equal to the sum of the first n terms of the A.P. series 57, 59, 61, … then what is the value of n?

Solution: The condition given in the question says that S2n = Sn

Hence, 2n/2 [2.2 + (2n-1).3] = n/2 [2.57 + (n-1) 2]

So, (4+6n-3) = 1/2 (114 +2n -2)

This gives 6n +1 = 57 + n - 1

Hence, n =11.

askIITians is an online portal that offers a platform wherein students can get answers to their queries on various topics like principles of arithmetic progressions, condition for three terms to be in A.P etc. from the IIT experts.

Related resources:

To read more, Buy study materials of Sequences and Series comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.

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