Sequence and Series Formula for JEE Main/Advanced

Sequence

It means an arrangement of number in definite order according to some rule

Important Points

The various numbers occurring in a sequence are called its terms. The number of terms is called the length of the series
Terms of a sequence are denoted generally by $a_1 , a_2, a_3, ….., a_n$ etc., The subscripts denote the position of the term. We can choose any other letter to denote it
The nth term is the number at the nth position of the sequence and is denoted by a_n
The nth term is also called the general term of the sequence and given by $t_n = S_n – S_{n-1}$

Arithmetic Progression

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant

a, a+ d, a+2d. a+ 3d

Important formula

1. How to Prove if the sequence is AP given nth term

if $T_n – T_{n-1}$ is independent of n and is constant

Generally if $T_n= An + B$ , then it will be in AP

2. nth Term of A.P from beginning

$T_n=a + (n-1)d = l$

3. nth Term of A.P from end

$T_n^{‘}=l – (n-1)d =a$

4. $T_n + T_n^{‘} = a + l$

This also proves that the sum of the terms equidistant from beginning and end are equal and is same as a +l

i.e if an A.P has 5 terms $a_1, a_2,a_3,a_4,a_5$,then

$a_1 + a_5= a_2 + a_4 = a_3 + a_3$

5. if a, b and c are in A.P, then

$2b= a+c$

6. if a and l are first and last term of the A.P of n items, then

$d= \frac {l-a}{n-1}$

6. Sum of the n terms of the A.P

$S_n=\frac {n}{2} [a +l]$

$S_n=\frac {n}{2} [2a + (n-1)d]$

$S_n=\frac {n}{2} [T_n+ T_n^{‘}]$

7. Some General Formulas Based on Above

$1 + 2 + 3 + 4…+ n= \frac {n(n+1)}{2}$

$1 + 3 + 5..+ 2n-1= n^2$

8. Some More Formula’s

$d= T_n – T_{n-1}$
$T_n= S_n-S_{n-1}$
$d= S_n – 2S_{n-1} + S_{n-2}$

9.How to Prove if the sequence is AP given $S_n$

if $S_n$ is of the form, $S_n= An^2 + Bn$ then it will be in AP

10. If we have to choose three members of A.P

a-d, a a +d

11. If we have to choose four members of A.P

a- 3d, a-d, a+ d, a+ 3d

12. If we have to choose five members of A.P

a- 2d, a-d, a, a+ d, a+ 2d

13. Arithmetic Mean is given by

$AM= \frac {a+b}{2}$

Geometric Progressions

A sequence is said to be a geometric progression or G.P., if the ratio of any term to its preceding term is same throughout.

$a, ar, ar^2 ,ar^3$

Common difference =r
$a \ne 0$ and $ r \ne 0$

Important formula

1.nth term of GP from Beginning

$T_n=ar^{n-1}=l$

2. nth term of GP from last

$T_n^{‘}=\frac {l}{r^{n-1}}=a$

3. $T_n \times T_n^{‘} = a \times l$

It also proves that Product of kth term from beginning and last is independent and is equal to $a \times l$

4. if a, b and c are in G.P
$b^2 = ac$

5. If a and l are first and last term of the G.P

$r= (\frac {l}{a})^{1/n-1}$

6. Sum of G.P

$S_n =a \frac {r^n -1}{r-1}$ if r >1

$S_n =a \frac {r^n -1}{r-1}$ if r >1
$S_n = na$ if r =1

7. if the G.P is infinite

if |r| < 1

then $S_n = \frac {a}{1-r}$

8. if $a_1, a_2,a_3 ..a_n$ are in G.P

then

$\frac {1}{a_1} , \frac {1}{a_2}, \frac {1}{a_3} …. , \frac {1}{a_n}$ are in G.P with common difference $\frac {1}{r}$

$a_1^n , a_2^n, a_3^n ..$ are in G.P with common difference $r^n$

Also

$a_1a_n= a_2a_{n-1} = a_3a_{n-2}=..$

9. if the three number of G.P are to be taken

$\frac {a}{r}, a , ar$

10. if the five number of G.P are to be taken

$\frac {a}{r^2},\frac {a}{r}, a , ar,ar^2$

11. If $a_1,a_2 ,a_3$ are in G.P

$log a_1 , log a_2, log a_3$ are in A.P

12. Geometric Mean is given by

$GM= \sqrt {ab}$

Formula based on Geometric Progression

(a) Recurring Decimal

R=.PQQQ….
let p be number of digits in P and q be the number of digits in Q
$10^p R =P.QQQQ…$
$10^{p+q} R = PQQQQQ…$

Subtracting above two

$R= \frac {PQ -P}{10^{p+q} – 10^p}$

Example
R=.32585858..
Then
$R= \frac {3258 -32}{10^4 -10^2} = \frac {3226}{9900}= \frac {1613}{4950}$

If R=.QQQQ
then
$R= \frac {Q}{10^q -1}$

(b) $S_n= a + aa + aaa + ….$ where $a \n N , 1 \leq a \leq 9$
example
$S_1 = 1 + 11 + 111 + 1111 +….$
$S_2 = 2 + 22 + 222..$
then
$S_n= \frac {a}{9} [ \frac {10}{9} (10^n -1) -n]$

Inequalities

A.M > G.M

$a+ b \geq 2 \sqrt {ab}$

$\frac {a_1 + a_2 + a_3 + …a_n}{n} \geq (a_1a_2a_3..a_n)^{1/n}$

Arithmetic Geometric Series

Let $S =a + (a+d)r + (a+2d)r^2 +….. + [a + (n–1)d]r^{n-1}$

Here we have both the AP and G.P in the terms. Lets see how to to calculate the sum

$rS= a r+ (a+d)r^2 + (a+2d)r^3 +….. + [a + (n-1)d]r^{n}$

Subtracting we get

$(1-r)S= a + (dr + dr^2 + ..dr^{n-1}) -[a + (n-1)d]r^{n}$
$1-r)S= a +\frac {dr(1-r^{n-1})}{1-r} -[a + (n-1)d]r^{n}$

$S= \frac {a}{1-r} +\frac {dr(1-r^{n-1})}{(1-r)^2} -\frac {[a + (n-1)d]r^{n}}{1-r}$

when |r| < 1. $n -> \infty$, $r^n=0$, hence

$S=frac {a}{1-r} +\frac {dr}{(1-r)^2}

Important Formula on Sigma

$\sum_{k=1}^{n} T_k= T_1 + T_2 + T_3 +…+T_n$

$\sum_{k=1}^{n} (T_k \pm T^{‘}_k=\sum_{k=1}^{n} T_k \pm \sum_{k=1}^{n} T_k^{‘}$

$\sum_{r=1}^{n} f(r+1) – f(r)= f(n+1) – f(1)$

$\sum_{r=1}^{n} f(r+2) – f(r)= f(n+2) + f(n+1) – f(2) -f(1)$

$\sum_{n=1}^{n} n = \frac {n(n+1)}{2}$

$\sum_{n=1}^{n} n^2 = \frac {n(n+1)(2n+1)}{6}$

$\sum_{n=1}^{n} n^3 = [\frac {n(n+1)}{2}]^2$

$\sum_{n=1}^{n} n^4 = \frac {n(n+1)(2n+1)(3n^2 + 3n -1)}{30}$

$\sum_{n=1}^{n} a^n =\frac {a(a^n -1)}{a-1}$

Rules for finding the sum of Series

a. Write the nth term T_nof the series

b. Write the T_n in the polynomial form of n

$T_n= a n^3 + bn^2 + cn +d$

c. The sum of series can be written as

$ \sum S = a \sum {n^3} + b \sum {n^2} + c \sum n + nd$

Method of Difference

Many times , nth term of the series can be determined. For example

5 + 11 + 19 + 29 + 41……

If the series is such that difference between successive terms are either in A.P or G.P, then we can the nth term using method of difference

$S_n = 5 + 11 + 19 + 29 + … + a_{n-1} + a_n$

or $S_n= 5 + 11 + 19 + … + + a_{n-1} + a_n$

On subtraction, we get

$0 = 5 + [6 + 8 + 10 + 12 + …(n – 1) terms] – a_n$

Here 6,8,10 is in A.P,So

$a_n = 5 + \frac {n-1}{2} [12 + (n-1)2]$

or $ a_n= n^2 + 3n + 1$

Now it is easy to find the Sum of the series

$S_n = \sum_{k=1}^{n} {k^2 +3k +1}$

$=\frac {n(n+2)(n+4)}{3}$

Short Cut Formula

If the First difference is in AP or GP

for AP

$T_n= a(n-1)(n-2) + b(n-1) + c$

we can find a,b,c using terms 1,2, 3

for GP

$T_n= ar^{n-1} + bn + c$

If the difference of the differences are in AP or GP