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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
a_{n} = a+(n-1)d.
The sum of first n terms is given by
S_{n} = n/2 [2a + (n-1) d]
Watch this Video for more reference
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 a_{1}, a_{2}, a_{3}, …, a_{n} are in A.P.
then a_{1} + k, a_{2 }+ k, a_{3} + k ……, a_{n} + k and
a_{1} –k, a_{2} – k, a_{3} – k, ……, a_{n}–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.
then ka_{1}, ka_{2}, ka_{3} ……, ka_{n} and
a_{1}/k, a_{2}/k, a_{3}/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 a_{1}, a_{2}, a_{3}, …, a_{n} and b_{1}, b_{2}, b_{3}, …, b_{n}, are in A.P.
then a_{1}+b_{1}, a_{2}+b_{2}, a_{3}+b_{3}, ……, a_{n}+b_{n}.
a_{1}–b_{1}, a_{2}–b_{2}, a_{3}–b_{3}, ……, a_{n}–b_{n} will also be in A.P.
4. Reversing the order of an A.P. also results an A.P.
e.g. Suppose a_{1}, a_{2}, a_{3}, ……, a_{n} are in A.P.
then a_{n}, a_{n–1}, ……, a_{3}, a_{2}, a_{1} will also be in A.P.
1. If nth term of an series is t_{n} = An + B, then
t_{n} - t_{n–1} = An + B – A(n–1)–B
= A = constant
i.e. the series is in A.P.
2. If t_{n} = An^{2} + Bn + c, then
t_{n} – t_{n–1} = A(n^{2}) + Bn + C – (A(n–1)^{2} + B(n–1) + c)
= A (2n + 1) + B = ∪(n) (say)
Then U_{n} – U_{n–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.
t_{n} = An^{3} + Bn^{2} + 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 a_{1}, a_{2}, a_{3}, …… and b_{1}, b_{2}, b_{3}, …… are two A.P.’s with common differences d and d’ respectively then a_{1} + b_{1}, a_{2} + b_{2}, + a_{3} + b_{3}, … is also an A.P. with common difference d + d’.
6. If a_{1}, a_{2}, a_{3}, ……, a_{n} are in A.P., then a_{1} + a_{n} = a_{2} + a_{n–1} = a_{3} + a_{n–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} + 9^{2 }- x^{2}]/2.10.9
This implies 1/2 = (181-x^{2})/ 180
So x^{2} = 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 S_{2n} = S_{n}’
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.
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