Sequence and Series is one of the most important chapter that is covered in the syllabus for IIT-JEE. The summation of series is one of the most important way to reduce large series into some value and it is widely used throughout the syllabus. The tricks and techniques discussed here are very useful and are widely used in other chapters as well. The A.M- G.M inequality which will be discussed here is also a very important result that is used widely.* *

So Let us Probe more into this Chapter...

**Arithmetic Progression (A.P): **
A sequence of numbers is said to be in A.P when difference _{ } is constant for all

n

ÃŽ N. This constant

_{ } is called common difference. If a the first term and d the common difference of the A.P then,

1) n^{th} term t_{n }= a+ (n-1) d.

Proof

1^{st} term of A.P = a = a+ (1-1) d.

2^{nd} term of A.P = a+d = a+ (2-1) d.

3^{rd} term of A.P = a+2d = a + (3-1) d.

4^{th} term of A.P = a+3d = a + (4-1) d.

\ n^{th} term of A.P = a + (n-1) d.

2) Sum of n terms

_{ }

Why?

Let S_{n} = a + (a+d) + (a+2d) + ------- + a+ (n-2) d + a+ (n-1) d --------- (1)

Or S_{n} = a+ (n-1) d + a+ (n-2) d + ------------ + (a+d) + a ---------- (2)

Now adding (1) and (2) we get

_{ }

**Arithmetic Means (A.M):**

The n number A_{1}, A_{2}, A_{3}---------, A_{n} are arithmetic means between a and b if a, A_{1}, A_{2}, -----A_{n}, b are in A.P.

Since a is first term and b is (n+2)^{ th} term.

_{ }

_{ }

__Note: __

1) _{ }

Why?

_{ }

2) Arithmetic mean A of n numbers a_{1}, a_{2}, ----------- a_{n} is given by,

_{ }

Illustration 1:

For what value of n _{ } is A.M between a and b?

A.M between a and b is _{ }

_{ }

**Properties of A.P: **

1) If a fixed number is added (substracted) to each term of a given A.P. then resulting sequence is also A.P with same common difference as given A.P.

Dumb Question:

1) Why common difference has not changed?

Ans: Suppose the original A.P had_{ } , now let g be substracted from each term. So_{ } . Therefore only the first term of A.P has changed.

2) If each term of an A.P is multiplied (or divided) by fixed constant then resulting sequence is also an A.P with common difference multiplied (or divided) by same constant.

3) Sum and difference of corresponding terms of two A.P’s will form an A.P.

Why?

Let first A.P be

_{ }

And second A.P be

_{ }

_{ }

So, resulting sequence is A.P with first term a_{1}+a_{2} and common difference d_{1}+d_{2}.

4) If we want to pick terms of an A.P then convenient way of doing that is,

For three term’s in A.P we choose a-d, a, a+d.

For four terms in A.P we choose a-3d, a-d, a+d, a+3d.

5) _{ }

6) If a, b, c are in A.P Ãž 2b = a+c or b-a = c-b.

7) The sum of the terms of an A.P equidistant from beginning and end is constant and equal to sum of first and last term.

_{ } = Constant = _{ } .

Illustration 2:

Find the number of terms in the series _{ } of which the sum is 300.

Solution:

Here we observe that a=20,_{ } , and S_{n} = 300.

_{ }

Dumb Question:

1) Why do we get n as 25 and 36 both?

Ans: This clearly indicates that sum of terms from n = 26 to 36 is zero.

i.e. _{ }

** **

**Geometric Progression:**

A sequence is a G.P when its first term is non-zero and each of its succeeding term is r times the preceding term. The fixed term r is known as common ratio of G.P.

If a is first term of an G.P and r its common ratio then,

1) n^{th} term_{ } .

Why?

Let us observe the pattern

_{ }

So, clearly_{ } .

2) The sum of first n terms

_{ }

Why?

Case 1: Suppose r=1

So, _{ }

= na

Case 2: If r ¹ 1.

_{ }

3) If |r|<1 and n® ¥ then _{ }

Why?

_{ }

Dumb Question:

1) Why _{ } r^{n}=0?

Ans: Here |r|<1 so as we multiply r by itself so the value goes down and down. For example. Suppose r = ½, so r^{2}=1/4 is less than ½ and so the value keeps going down.

Illustration 3:

If a_{1}, a_{2}, a_{3} are 3 consecutive terms of a G.P with common ratio k and a_{1}<0. Find values of k for which a_{3}> 4a_{2}-3a_{1} is satisfied.

Solution:

Now since a_{1}, a_{2}, a_{3} are in G.P

So a_{2} = a_{1}k.

a_{3} = a_{1}k^{2}.

\ a_{1}k^{2} > 4a_{1}k-3a_{1}.

Ãž a_{1} (k^{2}-4k+3) > 0.

Or a_{1} (k-1) (k-3) > 0.

Since a_{1}<0 so it means (k-1) (k-3) < 0.

Ãž 1<k<3.

**Geometric Means (G.M)**** **

If G_{1}, G_{2}, G_{3}, G_{4}---------G_{n} are in G.M’s between a and b then a, G_{1}, G_{2}, ----------- G_{n}, b are in G.P.

Now b is t_{n+2} so b = ar^{n+1}

_{ }

Note:

1) The product of n G.M’s between a and b is equal to n^{th} power of one G.M between a and b

i.e. G_{1}, G_{2}------- G_{n} = _{ } .

Why?

_{ }

2) If a_{1}, a_{2}, ------- a_{n} are n non-zero numbers then their G.M is given by

_{ }

Illustration 4:

If we insert odd number (2n+1) G.M’s between 4 and 2916 then find the value of (n+1)^{ th} G.M?

Solution:

Now 4, G_{1}, G_{2}, G_{3}----------- G_{n+1}, ------- G_{2n}, G_{2n+1}, 2916 are in G.P. So G_{n+1} will be the middle mean of (2n+1) odd means and so it will be equidistant from 1^{st} and last term.

So, 4, G_{n+1}, 2916 are also in G.P.

And thus,

_{ }

= 108.

**Properties of G.P: **

1) If each term of a G.P is multiplied (or divided) by some non-zero quantity the resulting progression is G.P with some common ratio.

Why?

Suppose the G.P is with _{ }

Now this G.P is multiplied by some k (¹0).

So G.P will be _{ }

So, only the first term of G.P has changed and the common ratio remains unaffected.

2) If a_{1}, a_{2}, -------- and b_{1}, b_{2}, ---------- be two G.P’s of common ratio r_{1} and r_{2} respectively, then a_{1}b_{1}, a_{2}b_{2}, ------ and _{ } will also form G.P common ratio will be r_{1}r_{2} and _{ } respectively.

Why?

Let the series a_{1}, a_{2}--------- have the n^{th} term as _{ } and the series b_{1}, b_{2}-------- have n^{th} term as_{ } .

So the series a_{1}b_{1}, a_{2}b_{2} ------- will have n^{th} term as

_{ }

So the common ratio now becomes r_{1}r_{2}.

3) If we have to take three terms in G.P we take them as _{ } with common ratio r and four terms as _{ } with common ratio r^{2}.

4) If a, b, c are in G.P then b^{2} = ac.

5) If a_{1}, a_{2}, a_{3} -------- a_{n} is a G.P then loga_{1}, loga_{2}--------- loga_{n} is an A.P.

Why?

Now a_{1}, a_{2}, a_{3} ------- a_{n} form a G.P.

So _{ }

_{ }

This is clearly term of an A.P.

Dumb Question:

1) How does loga_{1} + (i-1) logr represents term of an A.P?

Ans: Let loga_{1} = A

And logr = D

So, loga_{1}+ (i-1) logr = A+ (i-1) D.

So, A+ (i-1) D is term of an A.P with first term as A and common difference as D.