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# 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 an
• 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

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

or

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

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}$### Rules for finding the sum of Series a. Write the nth term Tnof the series b. Write the Tn in the polynomial form of n$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$If the difference of the differences are in AP or GP for AP$T_n= a(n-1)(n-2)(n-3) + b(n-1)(n-2) + c(n-1) + d$we can find a,b,c ,d using terms 1,2, 3,4 for GP$T_n= ar^{n-1} + bn^2 + cn +d\$

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