# Sequence

Sequence is an ordered collection of numbers.

For example, the numbers x1, x2, x3, x4, ……………., xn, is called a sequence, if according to some rule or law, there is a definite value of xn, (called the term or element of the sequence), corresponding to any value of the natural number  n.

x1 is the 1st term of the sequence , x2 is the 2nd term, …………………., xn is the nth term.

In the nth term xn, by putting n = 1,2,3,……. successively , we get x1, x2, x3, x4, ………

The nth term of a sequence is a function of the positive integer n. The nth term is also called the general term of the sequence. To specify a sequence, nth term must be known.

Ex. Determine the General Term of the sequence, $\displaystyle 1,\text{ }\frac{1}{4},\text{ }\frac{1}{9},\text{ }\frac{1}{{16}}\ldots .$

$\displaystyle {{A}_{1}}=\text{ }\frac{{\left( 1 \right)}}{{{{{\left( 1 \right)}}^{2}}}}$$\displaystyle ,~{{A}_{2}}=\frac{{\left( 1 \right)}}{{{{{\left( 2 \right)}}^{2}}}},$$\displaystyle {{A}_{3}}=\frac{{\left( 1 \right)}}{{{{{\left( 3 \right)}}^{2}}}}.$$\displaystyle So,\text{ }{{A}_{n}}=\frac{{\left( 1 \right)}}{{{{{\left( n \right)}}^{2}}}}.$

So, the General term $\displaystyle {{t}_{n}}=\frac{{\left( 1 \right)}}{{{{{\left( n \right)}}^{2}}}}.$

Finite Sequence : If the number of elements in a sequence is finite, the sequence is called finite

sequence;

A finite sequence x1, x2, x3, x4 ….. xn is denoted by {xi} n i=1

Examples of Finite Sequence

• A sequence of even positive integers within 15 i.e., is 2, 4, 6, 10, 12, 14.
• A sequence of odd positive integers within 8 i.e., is 1, 3, 5, 7, etc.

Infinite Sequence : If the number of elements is unending, the sequence is infinite.

Infinite sequence x1, x2, x3, x4 …..  is denoted by {xi} i=1

or simply by {xn} where xn is the nth element of the sequence.

Examples of Infinite Sequence

A sequence of even positive integers is 2, 4, 6,8, 10, 12,  …………………………………….

A sequence of odd positive integers is 1, 3, 5, 7, 9, 11, 13,   …………………………………

The sequence {1/n} is 1,1/2, 1/3…

The sequence (-1)n is -1, 2, -3, 4, -5….

# Series

Series is the sum of the terms of the sequence.

An expression of the form a1 + a2 + a3 + ……+ an + ………………….which is the sum of the elements of the sequence { an} is called a series.

If the series contains a finite number of elements, it is called a finite series, otherwise it is called an infinite series.

In the Series, Sn = u1 + u2 + u3 + u4 + ……..+ un,  Sn is called the sum to n terms (or the sum of the first n terms) of the series and is denoted by by åSn (sigma)

Sn=åur or r=1 n , which is simply written as  åun

Ex. :

• 1 + 5 + 9 + 13 + 17 ……………………. is a series in which 1st term = 1, 2nd term = 5, and so on.
• 2 − 4 + 8 − 16 + ……………….. is a series in which 1st  term = 2, 2nd term = −4, and so on.

The general term of a series is represented as tn. So,tn= s n -s n-1

Ex : For S n = n2 +2n find t n and then a 1 & a 2.

Step 1 : S n = n2 +2n. So, s n-1 = (n-1) 2 +2(n-1) = (n2  + 1 – 2n) + (2n – 2) = n2  – 1

Step  2 : t n = s n – s n -1 = [(n2 +2n )] – [(n2  – 1)] =  n2 +2n  – n2  + 1 =  2n + 1

Step 3 :  a 1= t 1 =2.(1) + 1 = 2 +1 =3, a 2 = t 2  = 2.(2) + 1 = 5.

# Arithmetic Progression

A series is said to be in Arithmetic Progression (A.P) if the consecutive numbers in the series differs by a constant value.

So, in an A.P series, , a1, a2, ..an,  (a2-a1)=(a3-a2)=(an– an-1)

The constant value is referred to as “common difference”. The series in which the consecutive terms increases by a constant quantity, is referred to as an increasing series and if terms decrease by a constant quantity it is referred to as a decreasing series.

( i) The series 4, 9, 14, 19, 24, 29 ………….. (increasing series, CD = 5)

( ii) The series 12, 8, 4, 0, -4 ………… (decreasing series, CD = -4)

In an A.P. the first number is denoted by “a” and the common difference is denoted by “d”. The consecutive terms of arithmetic progression are:

a, a+ d, a+2d, a+3d, ………..a+ nd

The first term is a, the second term is a+ d, the third term being a+ 2d. For the first term, the coefficient of d is zero, for the second term it is 1 and for the third term it is 2.

The nth term in AP is represented as : Tn = a + (n − 1)d

So, 1st term is a+(1-1)d = a+0=a

2nd  term is a+(2-1)d = a+d

3rd   term is a+(3-1)d = a+2d

…..

nth term is a + (n − 1)d

The last term in the series is denoted by I. So, I = a + (n − 1)d].

Sum of a Number of Terms in A.P.

The terms in an A.P. are given by

a, a +d, a +2d, a +3d, ………a + (n − 2)d, a + (n − 1)d

The sum of all these terms which is denoted by “S” is given by

S = (n/2) x {2a + (n − 1)d}

Since l = l = a + (n − 1)d, the above equation may also be written as

S = (n/2)x {a+a + (n − 1)d}
$\displaystyle =\left( {\frac{n}{2}} \right)\times \left( {a+1} \right),$where a is the first term and l is the last term

# Arithmetic Progression – Problems

Arithmetic Progression – Problems & Solutions

Ex.1 : If the first term of an A.P. ‘a’ = 5 and the common difference ‘d’ = 3. Find the first five term and the General Term (nth  term)

Step 1 : First 5 terms

T1 = a = 5, T2 = a + d = 5+3 = 8, T3 = a + 2d = 5+2.(3) = 11, T4 = a + 3d= 5+3.(3) = 14,

T5 = a + 4d= 5+4.(3) = 17

Step 2 : General Terms

Tn   = a + (n − 1)d = 5 + (n − 1).(3) = 5 + 3n − 3 = 3n +2

Ex.2: The 8th  term of an AP is 21 and 32nd  term is 45, determine the 20th term.

Step 1 : Here t8 = 21. T32= 45. Let a be the first term, d =common difference, n =number of terms

Step 2 :  t8 = a + (8 – 1 )d = a +7d, or a + 7d = 21  … (Eq 1)

Step 3 : t32 = a + (32 – 1 )d = a + 31d = 45 ….(Eq 2)

Step 4 : Subtracting Eq 1 from Eq 2, we get 24d= 24, or d=1

Step 5 : Substituting the value of d in Eq 1, we get  a+7d = 21, or a+7 = 21. So, a=14

Step 6: t20 = a + (20– 1 )d= 14+ 19×1 = 33

# Arithmetic Progression – Problems

Arithmetic Progression – Problems & Solutions

Ex.1 : The first term of A.P. is  25, 11th term is 325. Find common difference.

Step 1: a = 25,  t11 =325

Step 2: t11 = a + (11 – 1 )d = a + 10d = 325, or 25 + 10d = 325, or 10 d = 300, or d=30

So, common difference d=30

Ex.2: First term of a series is 5 and the last term is 41. The sum of the series is 230. Find the number of elements  and the common difference.

$\displaystyle S=\left( {\frac{n}{2}} \right)\times \left( {a+1} \right)$ where a =first term, l =last term, S= sum of the series, n=number of terms

$\displaystyle So,\text{ }230=\left( {\frac{n}{2}} \right)\times \left( {5+41} \right),$
or 23n = 230, or n=10

Now T10 =  5 +(10-1) x d =41, or 5+9d = 41, or 9d = 36, or d=4

# Arithmetic Progression – Problems

Arithmetic Progression – Problems & Solutions

Ex1. In an A.P. the 4th and 6th term are respectively 9 and 13. Find the sum of first 24 terms of that A.P.

Here, t4 = 9 and t6 =13

Step 1 : t4 = a+ (4-1).d = a +3d = 9 … (Eq 1)

Step 2 : t6 = a+ (6-1).d = a +5d = 13 … (Eq 2)

Step 3 : Subtracting Eq 1 from Eq 2, we get 2d=4, or d=2

Step 4 : Sn = [(n/2) x {2a +(n-1) d}]

Step  5 : Substituting the values of a, n & d, we get

S24 = (24/2) x {2×3 + (24-1) x2 } = 12 x (6+46) = 12 x 52= 624= 12 x (6+46) = 12 x 52= 624

Ex.2 : Which term of the series 9, 14, 19, ….. is 94.

Step 1 : Let the nth term of the series be 94. Here a = 9, d = 5 and tn =94

Step 2 : tn = a + (n – 1 )d. So, 94 = 9 + (n – 1 ) x 5, or 5(n – 1 ) = 94 – 9, or 5(n – 1 ) =  85, or n – 1 = 17, or n=18. Hence 18th term is 94

# Arithmetic Progression – Problems

Arithmetic Progression – Problems & Solutions

The sum of three term of an A.P. is 30 and their product is 750. Find them.

Let three terms of an A.P. be a – d, a, a + d. (We have taken these terms in this form as the computation would be much easier)

Step 1 : So, (a – d) + (a) + (a+d) =30  (because sum of the terms is 30)… (Eq 1)

And  (a – d) ´ a ´ (a +d) =750 (because sum of the terms is 750)… (Eq 2)

Step 2 :  Form Eq 1, we get, 3a=30. So, a=10

Form Eq (2), we get  a ´(a2 – d2)  = 750. Or a3 -ad2 =750. Or (10)3 – (10´d2)=750,

Or 1000 -10d2 =750,  or 10d2 = 250, or d2=25, or d=+ 5

Step 3 : So, the terms are a, a-d & a+d

Putting the value of d=5, we get the terms as 10-5, 10 & 10+5 , or 5, 10, 15

Putting the value of d=-5, we get the terms as 10- (-5), 10 & 10 + (-5) , or 15, 10, 5

# Arithmetic Mean

Arithmetic Mean – Problems & Solutions

When three quantities are in A.P., then the middle one is said to be the arithmetic mean of the other two.

If a, b and c are in A.P., then b is the arithmetic mean of a & c.

Since a, b and c are in A.P., b − a =  c −b = d, Now b − a =  c −b, or  2b = a + c, or b=(a+c) / 2

Ex. Insert 14 arithmetic means between 4 and 64.

When 14 terms are inserted, the total number of terms in the series = 16 [taking into account  the first  term 4,  the last term  64, and 14 new inserted terms). So, a=4, t16 = 64, n=16

Now t16 = a + (16-1)d, So, 4+15d = 64, or 15d=60, or $\displaystyle d=\frac{{60}}{{15}}=4$

So, the series would be 4,(4+4), (4+8) .. 64, or  4,8,12,…64

# Arithmetic Mean – Problems

Arithmetic Mean – Problems & Solutions

Ex.1 : The sum of three terms in an A.P., is  45 and their product is 3360. Find the terms.

Step 1  Let us assume the three terms be a − d, a and a + d, where a-d is the first  term, a is the middle term and a+d is the last term

Step 2 : The sum of these three terms = a − d + a + a + d = 45, or 3a=45, or a=15

Step 3 : The product of the terms is( a − d)(a)(a + d) = 3360

or (15 − d)(15)(15 + d) = 3360,  or , 15 X (225 − d2) = 3360, or (225 − d2)
$\displaystyle =\frac{{3360}}{{15}}=224$

d2 = 225 − 224 = 1, or d=+1

Step 4: Considering d=1. We get the  terms as a-d, a, a+d, i.e 15-1, 15, 15+1, i.e 14,15, 16 (the increasing series)

Considering d=-1. We get the  terms as a-d, a, a+d, i.e 15- (-1), 15, 15+ (-1), i.e 16,15, 14 (the decreasing series)

Ex.2: Insert 3 A.M between 5 and 29.

Step 1 : Let the AM’s be x1, x2 & x3 between the terms 5 & 29. So, the elements in A.P series are  5, x1, x2, x3 , 29. Here a=5, t5 = 29, n=5 (number of elements)

Step 2 : tn = a + (n – 1 )d. So, t5 = a + (n – 1 )d = 29, or 5 + (5 – 1)d = 29, or 5+4d=29,
$\displaystyle or\text{ }d=\frac{{24}}{4}=6$

Step 3 : x1 = a + d =5  + 6 = 11, x2 =a + 2d = 5 + 6 x 2 = 5 + 12 = 17, x3 = a + 3d = 5 + 6 x 3 = 5 + 18 = 23.

# Geometric Progression (G.P)

A series is said to be in Geometric Progression (G.P) if the consecutive numbers in the series have a constant ratio. This constant ratio is referred to as “common ratio”.

In geometric progression, first term denoted  by ‘a’, common ratio ‘r’. ‘r’ remains constant whenever the ratio of any two consecutive terms is computed.

The terms of a G.P. are a, ar, ar2, ar3, ar4, ……………, arn−1, where a is the first term, r is the common ratio and arn−1 is the nth term

So, t1   =   a,  t1   =   ar ….. , tn   =   arn-1

Common Ratio of a G.P : In a G.P. each term except the first bears a constant ratio to its preceding term. This constant ratio is called the common ratio of a G.P.

Common Ratio  r =   (Any term) / (Preceding term) =  tn / tn-1

General term of a G.P. : If ‘a’ is the first term and r the common ratio then the general term of a GP is tn = ar (n-1)

Sum of a Number of terms in G.P : S= [a(1 − rn)] / (1 − r), where a = first term of the series, r= common ratio, n=number of terms and S= Sum of the  terms of the series

Sn= [a(1 − rn)] / (1 − r), where r<1

Sn= [a(rn -1)] / (r – 1), where r>1

Ex. First term in G.P.is 4 and the common ratio r is 3. Find the first three terms and the nth term.

Here a=4,r=3. tn = arn −1 = 4X(3)n−1

# Geometric Progression – Problems

Geometric Progression – Problems & Solutions

Ex. 1: Compute the sum of the series 81, 54, 36, ……………..upto 13 terms.

Step 1: First term ‘a’ = 81, common ratio  r= 54/81 = 2/3, n=13

Step 2: S13 = [a(1 − rn)] / ((1 − r) = [(81)(1−(2/3)13)] / (1− (2/3)13 = [(81)X( 1- .005138)] / (1/3) = [81×0.99486)] / 1/3 = 80.58366 x 3 = 241.750 appx

Ex 2: Find the 6th and 10th term of the series 1, 2, 4, 8,………..

Step 1 : r= 2/1 = 4/2 =2, a=1, tn=arn −1

Step 2 : t6 = ar 6-1 =ar 5 = 1.(2) 5 = 32, t10=ar 10-1 =ar 9 =1.(2) 9 = 512

Ex. 3: In a G.P. the first term is  3, last term is 384 and sum of the series is 765. Find the common ratio and the number of terms.

Here,a = 3, l = 384, s n = 765

Step 1 : Sn = [a(1 − rn)] / ((1 − r) = (a-rl) / (1-r)

Or 765 = [{(3- rX384)} / (1-r)], or 765(1 – r) = 3 – 384r, or 765 – 765r = 3 – 384r, or 762 = 381r,
$\displaystyle or\text{ }r=\frac{{762}}{{381}}=2$

Step 2 :  l= ar(n1), where l is the last term.

So, 384 = 3.2(n1), or 2(n1)  = 128 = 27. So, n-1 =7, or n=8

# Geometric Progression – Problems

Geometric Progression – Problems & Solutions

Ex. The sum and product of 3 numbers in G.P. are 65 and 3375. Find the numbers.

Step 1 : Let  the  3 numbers  in the G.P be, $\displaystyle \frac{a}{r}$, a, ar (we have designed the terms like tht to simplify the computation of value)

Step 2 : $\displaystyle So,\left( {\frac{a}{r}} \right)\times \left( a \right)\times \left( {ar} \right)=3375$ (product of the numbers). Or a3 = 3375, or a= (3375) 1/3 = 15

Step 3 : (a/r)+ (a)+ (ar) = 65, (sum of the numbers), or $\displaystyle a\left[ {\left( {\frac{1}{r}} \right)+1+r} \right]=65,$
$\displaystyle or~15\times \left[ {\left( {\frac{1}{r}} \right)+1+r} \right]=65$

$\displaystyle or\left[ {\left( {\frac{1}{r}} \right)+1+r} \right]=\frac{{65}}{{15}}$, $\displaystyle or\text{ }\left[ {\left( {\frac{1}{r}} \right)+r} \right]=\left( {\frac{{65}}{{15}}/} \right)-1,$
or {(1+r2) / r} $\displaystyle =\frac{{50}}{{15}}=\frac{{10}}{3}$

Step 4 : From step 3 , we get,3´ (1+r2) = 10r, or 3 + 3r2 = 10r, or 3r2 – 10r + 3 = 0, 3r2 – 9r – r  + 3 = 0,

Or 3r( r-3) – 1(r-3) =0, or  ( r-3) ( 3r-3) =0

So,  ( r-3) = 0, or ( 3r-1) = 0, So, r=3, or  r=1/3

Step 5 : So, the numbers are : $\displaystyle \frac{a}{r}$, r, ar

When r=3 the numbers are 15/3, 15, 15×3, or 5, 15, 45

When r=1/3 the numbers are $\displaystyle \frac{{15}}{{\left( {\frac{1}{3}} \right)}}$, 15, 15 x $\displaystyle {\left( {\frac{1}{3}} \right)}$, or 45, 15, 5

# Geometric Mean

Geometric Mean – Problems & Solutions

When three values a, b and c are in G.P., “b” is called the Geometric Mean of the numbers a  & c

Common Ratio : In G.P., the ratio of $\displaystyle \left( {\frac{b}{a}} \right)$ or the ratio of $\displaystyle \left( {\frac{c}{b}} \right)$  (which are equal) is referred as common ratio.

$\displaystyle \left( {\frac{b}{a}} \right)=\left( {\frac{c}{b}} \right)$, or  b2 = ac, or $\displaystyle b=\text{ }\pm \sqrt{{\left( {ac} \right)}}$

Ex 1. Find G.M. of  4 and 36?

G.M. = ± √ac = ± √(3 x 36) = √144 =12.

Ex 2 : Find the geometric mean between the two numbers 5 and 125.

If c be the G.M between two numbers, then c2=5 X 25 = 625. So, c= Ö625 = 25

So, the G.M between the numbers 5 & 125 is 25

Ex 3: Insert 5 geometric means between 5 and 320.

Inserting 5 terms in between we have 7 terms in the Geometric Series.

Step 1 : Here, the first term a= 5, the last term (or 7th term) is 320

Taking into account the first and last terms 5 and 320, we have 7 terms in all.

Step 2 : The first term ‘a’ = 5 and the seventh term t7 = 320

t7 =  ar7-1 = ar 6 = 5.r 6 = 320, or 5.r 6 = 320, or r 6 = 320 / 5 = 64.

So, r= (64)1/6 = (26) 1/6 = 2

Step 3 : So, the terms in G.P are, a, ar, ar2,  ar3, ar4, ar5

or, 5, 5×2, 5×22, 5×23 , 5×24, 5×25 , 5×26

Or, 5, 10, 20, 40, 80, 160, 320

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