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Sequence

arithmetic sequence
arithmetic sequence

A Sequence is a list of things (usually numbers) that are in order. let us consider the following of numbers.

(i) 1, 4, 7, 10,.....

(ii) 20, 18, 16, 14,....

(iii) 1, 3, 9, 27, 81,....

(iv) 1, 2, 3, 4, ...

We observe that each term after the first term

(i) is formed by adding 3 to the preceding term;

(ii) is formed by subtracting 2 from the preceding term; 

(iii) is formed by multiplying the preceding term by 3; each term in

(iv) is formed by squaring the natural numbers 1, 2, 3, 4,.....

In all the above case, we see that set of number follow a certain rule and we can easily say what number will come next to given number. thus, the numbers come in succession in accordance with a certain rule or low. A succession of numbers formed and arranged in a definite order according to a certain definite rule is called a sequence. the successive number in a sequence are called its terms.

 

Series

series examples
series examples

A series is formed by adding or subtracting the successive term of a sequence. A series is finite or infinite according to as the number of terms added in the corresponding sequence is finite or infinite.

eg. 1 + 4 + 7 + 10 + .......... + 25 is a finite series and 2 + 4 + 6 + 8 + ........... is finite series.

The successive numbers forming the series are called the terms of the series and the successive terms are denoted by t1, t2, t3,....., t,which denotes the 1st, 2nd, 3rd, ...... nth term respectively. The nth term, tn, of a series, is called its general term. Thus, in a series, 1+4+7+10+.......+25, the first term is 1, the second term is 4, the third term is 7, and so on.

Progression

Progression

A sequence of number is said to be a progression if the difference or ratio between its two successive terms is constant throughout the whole sequence. An example of progression is as follows.

(i) 1, 3, 5, 7,..... (ii) 1, 3, 9, 27,.....

In (i), the difference between two successive terms is equal to 2.

In (ii), the ratio of two successive terms is equal to 3.

Types of Progression

Progression is divided into following two types.

(i) Arithmetic progression

(ii) Geometric progression

Arithmetic progression or Sequence

A sequence is called an arithmetic progression if the difference between its two successive terms is constant throughout the whole sequence. An arithmetic progression can be denoted by A.P. The constant number obtained by subtracting succeeding term from its preceding term is called the common difference.

For example:-

(i) 1, 3, 5, 7, 9,....

(ii) 15, 12, 9, 6,......

From (i), we find that

second term - first term = 3 -1 = 2,

third term - second term = 5 - 3 = 2,

fourth term - third term = 7 - 5 = 2 and so on.

From (ii), we find that

second term - first term = 12-15 = -3,

third term - second term = 9 - 12 = -3,

fourth term - third term = 6 - 9 = -3 and so on.

Hence, the common difference 'd' is calculated by 

d =  succeeding term - proceeding term = tn - tn-1

Here, we find that the difference between two successive terms, in both sequences, are same or constant. So, such sequence is called arithmetic progressions. The C.D. of the two progressions are 2 and -3 respectively. Thus, arithmetic progressions is a series in which the successive terms increase or decrease by the common difference.

General term or nth term of an A.P.

To find the nth term of an A.P.

Let, t1 be the first term, n be the number of terms and 'd' the common difference of an A.P. respectively. Then,

t= a = a + (1-1)d

t= a + d = a + (2-1)d

t= a + 2d = a + (3-1)d

t= a + 3d = a + (4-1)d

In general, tn = a + (n-1)d

Formula: If tdenotes the nth term, of the arithmetic progression whose first term, common term and number of terms are a, d and n respectively.

With this term, arithmetic sequence and series can be written as:

Arithmetic sequence: a, a+d, a+2d, a+3d, ............

Arithmetic series: a+ (a+d) + (a+2d) + (a+3d), ..........

Arithmetic Mean

The terms between the arithmetic progression are known as arithmetic mean. Such as the three numbers 2, 4, 6 are in arithmetic progression with the common difference d = 2, then 4 is the arithmetic mean between 2 and 6.

For example:

Let a, b,c are in arithmetic progression

 b-a = c-b

or, b+b = a+c

or, 2b = a+c

or, b = \(\frac{a+c}{2}\)

Hence the arithmetic mean between a and c is (\(\frac{a+c}{2}\))

n Arithmetic Means between two numbers a and b

Let m1, m2, m3, .........mn be the arithmetic means between the given term a and b. Then, a, m1, m2, m3, .........mn, b are in A.P.

Here, numbers of arithmetic means = n

So, numbers of terms of A.P. = n+2

It means, 

    b = (n+2)th term of AP

or, b = a + (n+2-1)d, where d is common difference

or, b =a + (n+1)d

or, (n+1)d = b-a

∴ d= \(\frac{b-a}{n+1}\)

Now, m1 = a+d = a +  \(\frac{b-a}{n+1}\)

m2 = a + 2d = a +  \(\frac{2(b-a)}{n+1}\)

m3 =  a +  \(\frac{3(b-a)}{n+1}\)

.............................................

mn =  a +  \(\frac{n(b-a)}{n+1}\)

Sum of n terms of series in A.P.

Let us consider an arithmetic series

a + (a+d) + (a+2d) + (a+3d) + ...... + (l-2d) +(l-d) + l

Here, the first term = a, 

first term = a, 

common difference = d,

number of terms=  n,

last term (tn) = l

the term before last term = l-d

if the sum of n terms is denoted by Sn, then

Sn = a + (a+d) + (a+2d) + (a+3d) + ...... + (l-2d) +(l-d) + l .... (i)

Writing term in the reverse order,

Sn = l + (l-d) + (l-2d)  + ...... + (a+3d) + (a+2d) + (a+d) + a .... (ii) 

Adding the corresponding terms of (i) and (ii)

\(\frac{S_n \;= \;a \;+\; (a+d)\; +\; (a+2d)\; +\; (a+3d)\; + \;......\; + \;(l-2d) \;+\; (l-d)\; + \;l\\S_n\;=\;l\; +\;(l-d)\;+\;(l-2d)\;+\;......\;+\;(a+3d)\;+\;(a+2d)\;+\;(a+d)\;+a\:}{2S_n\;= \;(a+l) \;+ \;(a+l)\; + \;(a+l)\; +\; ............ \;+\; (a+l)\; + \;(a+l)\; +\; (a+l)}\)

=   n times (a+l)

=   n (a+l)

= \(\frac{n}{2}\)(a+l)

But, the last term l = a + (n-1)d

So, Sn = \(\frac{n}{2}\)(a+l) = \(\frac{n}{2}\)[a+a+(n-1)d] = \(\frac{n}{2}\)[2a+(n-1)d]

∴ Sn =  \(\frac{n}{2}\)[2a+(n-1)d]

Thus, if d is unknown, Sn = \(\frac{n}{2}\)(a+l)

And, if l is unknown, Sn = \(\frac{n}{2}\)[2a+(n-1)d]

1. Sum of first n natural numbers

the numbers 1, 2, 3, 4, ......, n are called the first n natural numbers. 

Here, first term (a) = 1

Common difference (d) = 2-1 = 1

Number of terms (n) = n

If Sn denotes the sum of these first n natural numbers, then

Sn = \(\frac{n}{2}\)[2a+(n-1)d] = \(\frac{n}{2}\)[2.1+(n-1).1] =  \(\frac{n}{2}\)[2+n-1] = \(\frac{n}{2}\)(n+1)

2. Sum of first n odd numbers

1, 3, 5, 7, ......., (2n-1) are the first n odd numbers.

Here, first term (a) = 1

Common difference (d) = 3-1 = 2

Number of terms (n) = n

If Sn denotes the sum of these first n odd numbers, then

Sn = \(\frac{n}{2}\)[2a+(n-1)d] = \(\frac{n}{2}\)[2.1+(n-1).2] =  \(\frac{n}{2}\)(2+2n-2) =  \(\frac{n}{2}\) × 2n = n2

3. Sum of first n even numbers

2, 4, 6, 8, ......., 2n are the first n even numbers.

Here, first term (a) = 2

Common difference (d) = 4-2 = 2

Number of terms (n) = n

If Sn denotes the sum of these first n even numbers, then

Sn = \(\frac{n}{2}\)[2a+(n-1)d] = \(\frac{n}{2}\)[2.2+(n-1).2] =  \(\frac{n}{2}\)(4+2n-2) =  \(\frac{n}{2}\)(2n-2) = n(n+1)

Geometric Progression or Sequence
Geometric Progression

In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3. Similarly,10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2.

Examples of a geometric sequence are powers rk of a fixed number r, such as 2and 3k. The general form of a geometric sequence is

a, ar, ar2, ar3, ............arn

where r ≠ 0 is the common ratio and a is a scale factor equal to the sequence's start value.

General Term or nth term of G.P.

We use the following notations for terms and expression involved in a geometrical progression:

The first term = a the nth term = toor b

The number of terms = n Common ratio = r.

The expression arn-1 gives us the nth term or the last term of the geometric progression whose first term, common ratio and a number of terms are a, r and n respectively.

∴tn = arn-1

With the help of this general term, geometric sequence and series can be written in the following ways:

Geometric Sequence: a, ar, ar2, ar\(^3\), .....

Geometric series: a + ar + ar2 + ar\(^3\) + ........

Geometric Mean

If the three numbers are in G.P., then the middle term is called the geometric mean of the other two terms. In other words, the geometric mean of two non-zero numbers is defined as the square root of their product.

Let a, G, b be three numbers in G. P., then the common ratio is the same i.e.

\(\frac{G}{a}\) =\(\frac{b}{G}\)

or, G2 = ab

or, G =\(\sqrt{a}{b}\)

Hence, the geometric mean of two numbers a and b is the square root of their product i.e. \(\sqrt{a}{b}\).

So, the geometric mean between two number 2 and 8 is G =\(\sqrt ab\) = \(\sqrt2*8\) = \(\sqrt16\) = 4.When

When any number of quantities are in G. P., all the terms in between the first and last terms are called the geometric means between these two quantities.

Here, Gn = arn = a \(\begin{pmatrix}b\\a\\ \end{pmatrix}\)\(\frac{n}{n + 1}\)

Relation between arithmetic mean and geometric mean

Geometric Means

"Arithmetic mean (A. M) is always greater than Geometric mean (G. M.) between two position real unequal numbers".

Let us consider two numbers 2 and 8

Here, AM between 2 and 8 =\(\frac{2 + 8}{2}\) = 5

GM between 2 and 8 = \(\sqrt 2 * 8\)) = 4

∴ AM > GM.

The sum of n terms of a series in G. P.

Let us consider geometric series a + ar + ar2 + ar\(^3\) + .......+ arn -3+ arni2+ arn-1

Here, first = a common ratio = r number of terms = n last term (l) = arn-1

∴ Sn = \(\frac{lr - a}{r - 1}\)

If the number of terms is odd, we take the middle term as aand the common ratio as r. If the number of terms is even, we take \(\frac{a}{r}\) and ar as the middle terms and r2 as the common ratio.

A series is formed by adding or subtracting the successive term of a sequence. A series is finite or infinite according to as the number of terms added in the corresponding sequence is finite or infinite.

eg. 1 + 4 + 7 + 10 + .......... + 25 is a finite series and 2 + 4 + 6 + 8 + ........... is finite series.

The successive numbers forming the series are called the terms of the series and the successive terms are denoted by$$ t_1, t_2, t_3,....., t_n,$$ where$$ t_1, t-2, t-3, ........ tn$$ denote the$$ 1^st, 2^nd, 3^rd ,...... .n^{nt}$$ term respectively. The n^th term, t_n, of a series, is called its general term. thus, in a series, 1+4+7+10+.......+25, the first term is 1, the second term is 4, the third term is 7, and so on.

The terms  between the first term and last term of an A.P are called arithmetic mean.,

.

Very Short Questions

\(\sum_{n=2}^5 (n^2- 3)\)

Putting n = 2, 3, 4, 5

= (22 - 3) + (32 - 3) + (42 - 3) + (52 - 3)

= (4 - 3) + (9 - 3) + (16 - 3) + (25 - 3)

= 1 + 6 +13 +22

= 42 Ans

un + 1 = 1 - \(\frac {1}{u_n}\) and u1 = 3, u2 and u3 = ?

If n = 1,un + 1 = 1 - \(\frac {1}{u_n}\)

u1 + 1 =1 - \(\frac {1}{u_1}\)

or, u2 = 1 - \(\frac 13\) = \(\frac {3 - 1}{3}\) = \(\frac 23\)

If n = 2, un + 1 =1 - \(\frac {1}{u_n}\)

u2 + 1 =1 - \(\frac {1}{u_2}\)

or, u3 = 1 - \(\cfrac{1}{\cfrac{2}{3}}\) = 1 - \(\frac 32\) = \(\frac {2 - 3}{2}\) = -\(\frac 12\)

∴ u2 = \(\frac 23\) and u3 = -\(\frac 12\) Ans

Sequence: A sequence is a set of numbers or quantities, which are formed according to some governed laws.

i.e. 1, 4, 16 ............................ and 2, 3, 4, 5...................

Series: The sum of the term of a sequence is called a series.

i.e. 1 + 4 + 16 + ....................... and 2 + 3 + 4 + 5 + ......................

Here,

\(\sum_{n=4}^7 (3n- 2)\)

where: n = 4, 5, 6, 7

\(\sum_{n=4}^7 (3n- 2)\)

= (3×4-2) + (3×5-2) +(3×6-2) +(3×7-2)

= (12-2) + (15-2) + (18-2) + (21-2)

= 10 + 13 + 16 + 19

= 58 Ans

\(\sum_{k=3}^7 (k^2 + 1)\)

= (32 + 1) +(42 + 1) +(52 + 1) +(62 + 1) +(72 + 1)

= (9 + 1) +(16 + 1) +(25 + 1) +(36 + 1) +(49 + 1)

= 10 + 17 + 26 + 37 + 50

= 140 Ans

Here,

First term (a) = 1

Second term (b) = 4

Common Difference (d) = b - a = 4 - 1 = 3

Last term (l) = 34

We know,

last term (l) = a + (n - 1)d

or, 34 = 1 + (n - 1) 3

or, 34 = 1 + 3n - 3

or, 3n - 2 = 34

or, 3n = 34 + 2

or, n = \(\frac {36}{3}\)

∴ n = 12

Again,

Sn = \(\frac n2\) [a + l]

Sn = \(\frac{12}{2}\) [1 + 34] = 6× 35 = 210

∴ Sn = 210 Ans

Here,

first term (a) = 2

second term (b) = 7

common difference (d) = b - a = 7 - 2 = 5

number of items (n) = 16

sum (S16) = ?

We know that,

Sn = \(\frac n2\) [2a + (n - 1) d]

S16 = \(\frac {16}{2}\) [2 × 2 + (16 - 1) 5] = 8 [4 + 15× 5] = 8 [4 + 75] = 8 × 79 = 632 Ans

Here,

first term (a) =3\(\frac12\) = \(\frac 72\)

second terrn (b) = 1

common difference (d) = b - a = 1 - \(\frac 72\) = \(\frac {2 - 7}{2}\) = -\(\frac 52\)

no. of items (n) = ?

last term (l) =-21\(\frac 12\) = -\(\frac {43}{2}\)

We know that,

l = a + (n - 1) d

or,-\(\frac {43}{2}\) = \(\frac 72\) + (n - 1) (-\(\frac 52\))

or, - 43 = 7 - 5n + 5

or, 5n = 12 + 43

or, n = \(\frac {55}{5}\) = 11

∴-21\(\frac 12\) is the 11th term. Ans

Here,

first term (a) = 20

common difference (d) = -3

numbers of term (n) = 7

seventh term (t7) = ?

We know that,

tn= a + (n - 1) d

t7 = 20 + (7 - 1) (-3) = 20 + 6 (-3) = 20 - 18 = 2 Ans

Here,

first term (a) = 2

second term (b) = 4

common difference (d) = b - a = 4 - 2 = 2

no. of terms (n) = 20

sum (S20) = ?

We know that,

Sn = \(\frac n2\) [2a + (n - 1) d]

S20 = \(\frac {20}{2}\) [2× 2 + (20 - 1)× 2] = 10 [4 + 38] = 10× 42 = 420 Ans

Here,

first term (a) = 1

second term (b) = 3

common difference (d) = b - a = 3 - 1 = 2

number of items (n) = 20

twenty term (t20) = ?

We know,

tn= a + (n - 1) d

t20 = 1 + (20 - 1) 2

t20 = 1 + 19× 2 = 1 + 38 = 39 Ans

Here,

first term (a) = 25

second term (b) = \(\frac {45}{2}\)

common difference (d) = b - a = \(\frac {45}{2}\) - 25 = \(\frac {45 - 50}{2}\) = -\(\frac 52\)

last term (l) = - 15

number of terms (n) = ?

We know that,

l = a + (n - 1) d

or, - 15 = 25 + (n - 1) - \(\frac 52\)

or, - 15 - 25 = -\(\frac 52\)n + \(\frac 52\)

or, - 40 - \(\frac 52\) = -\(\frac 52\)n

or - \(\frac {80 - 5}{2}\) = -\(\frac {5n}{2}\)

or, -\(\frac {85}{2}\)× -\(\frac {2}{5}\) = n

∴ n = 17

∴ The number of term = 17. Ans

Here,

first term (a) = 3

last term (l) = -9

number of terms (n) = 4 [including a and b]

common difference (d) = \(\frac {b - a}{n - 1}\) = \(\frac {-9 - 3}{4 - 1}\) = -\(\frac {12}{3}\) = - 4

x = a + d = 3 - 4 = - 1

y = a + 2d = 3 + 2× -4 = 3 - 8 = - 5

∴ x = -1 and y = -5 Ans

Here,

first term (a) = 2

second term (b) = - 9

common difference (d) = b - a = -9 - 2 = -11

last term (l) = -130

number of terms (n) = ?

We know,

l = a + (n - 1) d

or, -130 = 2 + (n - 1) (-11)

or, -130 - 2 = - 11n + 11

or, -132 - 11 = -11n

or, -11n = -143

or, n = \(\frac {-143}{-11}\)

∴ n = 13

Again, Sn = \(\frac n2\) (a + 1) = \(\frac {13}{2}\) (2 - 130) = 6.5× -128

∴ Sn = -832 Ans

Here,

common difference (d) = -3

number of terms (n) = 7

sum (S7) = 0

first term (a) = ?

We know,

Sn = \(\frac n2\) [2a + (n - 1) d]

or, 0 = \(\frac 72\) [2a + (7 - 1) (-3)]

or, 0× \(\frac 27\) = 2a + 6× (-3)

or, 0 = 2a - 18

or, 2a = 18

or, a = \(\frac {18}{2}\) = 9

∴ first term (a) = 9 Ans

Here,

first term (a) = \(\frac 14\)

second term (b) = \(\frac 12\)

common ratio (r) = \(\frac ba\) =\(\cfrac{\frac{1}{2}}{\cfrac{1}{4}}\) = \(\frac 12\)× \(\frac 41\) = 2

last term (l) = 128

number of terms (n) = ?

We know that,

l = arn-1

or, 128 = \(\frac 14\) (2)n-1

or, 27× 22 = 2n-1

or, 2n-1 = 29

or, n -1 = 9

or, n = 9 + 1

∴ n = 10

∴ The number of term = 10 Ans

Here,

first term (a) = 16

last term (l) = 36

We know that,

geometric mean (G.M.) =\(\sqrt {ab}\)

G.M. =\(\sqrt{({16}×{36})}\)

= \(\sqrt {({4^2}×{6^2})}\)

= \(\sqrt {({4×6})^2}\)

= \(\sqrt {({24})^2}\)

∴ Geometric mean = 24 Ans

Here,

first term (a) = \(\frac 13\)

second term (b) = 1

common ratio (r) = \(\frac ba\) = \(\cfrac {1}{\cfrac 13}\) = 3number of terms (n) = 8

number of terms (n) = 8

eighth term (t8) = ?

Using formula,

tn = arn-1

t8 = (\(\frac 13\)) (3)8-1 = \(\frac 13\)× 3× 36 = 729 Ans

Here,

first term (a) = 3

second term (b) = 1

common ratio (r) = \(\frac ba\) = -\(\frac 13\)

last term (l) = -\(\frac {1}{81}\)

number of terms (n) = ?

We know that,

l = arn-1

or, -\(\frac {1}{81}\) = 3 (-\(\frac 13 \))n-1

or, - \(\frac {1}{3^4× 3}\) = (-\(\frac 13\))n-1

or, (-\(\frac 13\))5 =(-\(\frac 13\))n-1

or, n - 1 = 5

or, n = 5 + 1

∴ n = 6

∴ The number of term = 6 Ans

Here,

first term (a) = \(\frac 23\)

second term (b) = -\(\frac 16\)

common ratio (r) =\(\frac ba\) = \(\cfrac{\frac{-1}{6}}{\cfrac{2}{3}}\)= -\(\frac 16\) × \(\frac 32\) = -\(\frac 14\)

number of items (n) = 5

5th term (t5) = ?

We know that,

tn= arn-1

or, t5 = \(\frac 23\)(-\(\frac 14\))5-1

or, t5 = \(\frac 23\)×(-\(\frac 14\))4

or, t5 = \(\frac 23\)× \(\frac {1}{16}\)×\(\frac {1}{16}\) = \(\frac {1}{384}\)

∴ fifth term = \(\frac {1}{384}\) Ans

Here,

first term (a) = \(\frac 23\)

second term (b) = x

third term (c) = \(\frac 32\)

We know that,

G.M. = \(\sqrt {ab}\)

x = \(\sqrt {(\frac 23) × (\frac 32)}\)

x = 1

r = \(\frac ba\) =\(\cfrac{1}{\cfrac{2}{3}}\) = \(\frac 32\)

y = ar = \(\frac 32\)× \(\frac 32\) = \(\frac 94\)

∴ x = 1 and y = \(\frac 94\) Ans

Here,

first term (a) = 12

second term (b) = 4 common ratio (r) = \(\frac ba\) = \(\frac 42\) = 2

number of terms (n) = 10

sum of the ten terms (S10) = ?

We know that,

Sn = \(\frac {a(r^n - 1)}{r - 1}\) = \(\frac {2(2^10 - 1)}{2 - 1}\) =\(\frac {2(1024 - 1)}{1}\) = 2× 1023 = 2046

∴ The sum of tenth terms = 2046 Ans

Here,

first term (a) = 81

second term (b) = 27

common ratio (r) = \(\frac ba\) = \(\frac {27}{81}\) = \(\frac 13\)

number of terms (n) = 5

sum of five terms (S5) = ?

We know,

Sn = \(\frac {a(r^n - 1)}{r - 1}\)

S5 = \(\frac {81[(\frac 13)^5 - 1]}{\frac 13 - 1}\)

=\(\frac {81[(\frac {1}{243}) - 1]}{\frac {1 - 3}{3}}\)

=\(\frac {81 × (\frac {1 - 243}{243})}{\frac {-2}{3}}\)

= 81× -\(\frac {242}{243}\)× -\(\frac 32\)

= 121

∴ The sum of five terms = 121 Ans

Let,

first term = a

common ratio = r

From equation,

S6 = 9 S3

or,\(\frac {a(r^6 - 1)}{r - 1}\) = 9 [\(\frac {a(r^3 - 1)}{r - 1}\)] [\(\because\)Sn =\(\frac {a(r^n - 1)}{r - 1}\)]

or, \(\frac {a[(r^3)^2 - (1)^2]}{r - 1}\)×\(\frac {r - 1}{a(r^3 - 1)}\) = 9

or, \(\frac{(r^3 + 1)(r^3 - 1)}{r^3 - 1}\) = 9

or, r3 = 9 - 1

or, r3 = 8

or, r3 = 23

∴ r = 2

∴ The common ratio = 2 Ans

Let: the three terms of an G.P. are \(\frac ar\), a and ar

From Question,

\(\frac ar\)× a× r = 729

or, a3 = 93

∴ a = 9

∴ The second term = 9 Ans

Here,

first term (a) = 6

third term (b) = x

A.M. = 30

A.M. = \(\frac {a + b}{2}\)

or, 30 = \(\frac {6 + x}{2}\)

or, 60 - 6 = x

∴ x = 54

G.M. = \(\sqrt {ab}\)

G.M. = \(\sqrt {6 × 54}\)

G.M. = \(\sqrt {2^3 × 3^3 ×3^3}\)

G.M. = 18

∴ The G.M. = 18 Ans

Here,

first term (a) = \(\frac 19\)

last term (l) = 9

number of Geometric mean (n) = 3

total number of term (n) = 5

3 Geometric mean (m1), (m2), (m3) = ?

l = arn-1

or, 9 = \(\frac 19\)r5-1

or, 81 = r4

or, 34 = r4

∴ r = 3

m1 = ar = \(\frac 19\)× 3 = \(\frac 13\)

m2 = ar2 = \(\frac 19\)× 32 = \(\frac 19\)× 9 = 1

m3 = ar3 = \(\frac 19\)× 33 = \(\frac 19\)× 27 = 3

∴ m1 = \(\frac 13\), m2 = 1, m3 = 3 Ans

If M and G be the arithmetic mean and geometric mean between two positive numbers a and b, then

m = \(\frac {a + b}{2}\) .................... (1)

G = \(\sqrt {ab}\) ................................(2)

Equation (1) - (2) we get

M - G = \(\frac {a + b}{2}\) - \(\sqrt {ab}\)

= \(\frac {a + b - 2\sqrt{ab}}{2}\)

= \(\frac {(\sqrt a)^2 - 2\sqrt a ⋅ \sqrt b + (\sqrt b)^2}{2}\)

= \(\frac 12\) (\(\sqrt a\) - \(\sqrt b\))2

\(\sqrt a\) - \(\sqrt b\) gives positive value

∴ M - G = \(\frac {(\sqrt a - \sqrt b)^2}{2}\)≥ 0

M - G≥ 0 or, M ≥G

∴ A.M.≥ G.M. Ans

Let:

Sn = 1, 3, 5, ..................... n

First term (a) = 1

Second term (b) = 3

Common difference (d) = b - a = 3 - 1 =2

number of items = n

We know,

Sn = \(\frac n2\)[2a + (n - 1) d]

or, Sn = \(\frac n2\)[ 2× 1 + (n - 1) 2]

or, Sn = \(\frac n2\)[2 + 2n - 2]

or, Sn = \(\frac n2\)× 2n

∴ Sn = n2Ans

Here,

first term (a) = 1

common ratio (r) = 2

sum (Sn) = 255

We know,S

Sn = \(\frac{a(r^n - 1)}{r - 1}\)

or, Sn =\(\frac{1(2^n - 1)}{2 - 1}\)

or,\(\frac{2^n - 1}{1}\) = 255

or, 2n = 255 + 1

or, 2n = 256

or, 2n = 28

∴ n = 8

∴ The no. of terms = 8 Ans

Let:

first term (a) = 3

last term (b) = 243

no. of geometric means (n) = 3

common ratio (r) = ?

3 G.M are X1, X2, X3 = ?

r = (\(\frac ba\))\(\frac {1}{n+1}\)= (\(\frac {243}{3}\))\(\frac {1}{3+1}\)= (81)\(\frac 14\)= (3)\(\frac {4×1}{4}\) = 3

X1 = ar = 3× 3 = 9

X2 = ar2 =3× (3)2 = 27

x3 = ar3 =3× (3)3 = 81

∴ 9, 27, 81 Ans

Let:

first term (a) = 27

last term (l) = \(\frac {32}{9}\)

number of terms (n) = 4

common ratio (r) = (\(\frac ba\))\(\frac {1}{n+1}\)

= (\(\cfrac{\frac{32}{9}}{27}\))\(\frac {1}{4+1}\)

=( \(\frac {32}{9×27}\))\(\frac 15\)

= (\(\frac 23\))\(\frac {5×1}{5}\)

=\(\frac 23\)

p = ar = 27×\(\frac 23\) = 18

q = ar2=27×\(\frac 23\)×\(\frac 23\) = 12

r = ar3 =27×\(\frac 23\)×\(\frac 23\)×\(\frac 23\) = 8

s = ar4 =27×\(\frac 23\)×\(\frac 23\)×\(\frac 23\)×\(\frac 23\) = \(\frac {16}{3}\)

∴ p = 18, q = 12, r = 8 and s = \(\frac {16}{3}\) Ans

Here,

first term (a) = 2

second term (b) = -6

common ratio (r) = \(\frac ba\) = \(\frac {-6}{2}\) = -3

number of terms (n) = 6

We know that,

Sn = \(\frac {a(1 - r^n)}{1 - r}\)

Sn = \(\frac {2(1 - (-3)^6)}{1 - (-3)}\)

Sn = \(\frac {2(1 -729)}{1 + 3}\)

Sn = \(\frac {2× (-729))}{4}\)

∴Sn= -364.5Ans

Here,

first term (a) = 3

second term (b) = 6

common ratio (r) = \(\frac ba\) =\(\frac 63\) = 2

last term (l) = 1535

sum of the series (Sn) = ?

Using formula,

Sn = \(\frac{lr - a}{r - 1}\) = \(\frac {1535 × 2 - 3}{2 - 1}\) = \(\frac {3070 - 3}{1}\) = 3067

∴ Sum (Sn) = 3067 Ans

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  • If f = {(1,2) ,  (2,3) , (4,5)} ,  find f (^-1) and the domain and range of f (^-1) .

    f (^-1) = {( 1,1) , (3,3) , (5,4)}
    Domain  of f (^-1) = {(2 ,3, 3)}
    Range of f (^-1) = {1,2,3}


    f (^-1) = {( 2,1) , (1,2) , (3,4)}
    Domain  of f (^-1) = {(5 ,3, 5)}
    Range of f (^-1) = {1,3,4}


    f (^-1) = {( 2,1) , (3,2) , (5,4)}
    Domain  of f (^-1) = {(2 ,3, 5)}
    Range of f (^-1) = {1,2,4}


    f (^-1) = {( 2,3) , (1,2) , (5,4)}
    Domain  of f (^-1) = {5,3,2)}
    Range of f (^-1) = {4,2,1}


  • If f   = {(1,4) , (2,4) , (3,5)} and g = {(4,2) , (5,2) }, then show the function gof in arrow diagram and find it ordered pair form.

    {(2,2) , (3,2) , (4,2)}


    {(1,2) , (2,2) , (3,2)}


    {(3,2) , (2,2) , (0,1)}


    {(3,2) , (2,2) , (0,2)}


  • If f = {(b,q) , (a,p) , (c,r)} and gof =  {(a,1) , (c,3) , (b,2)} , find the function g in terms of ordered pairs .

     

    {(p,1) , (q,2) , (r,3)}


    {(p,1) , (q,q) , (r,r)}


    {(1,1) , (q,2) , (q,3)}


    {(p,q) , (1,2) , (r,3)}


  • If f  is an identity function and g= {(1,3) , (2,4) , (3,5)} then write the composite function gof  in ordered pair form .

    {(1,3) , (2,4) ,(3,5)}


    {(1,2) , (2,4) ,(4,5)}


    {(1) , (2) ,(3,5)}


    {(1,2) , (2,4) ,(4,5)}


  • If gof = {(2,9) , (3,13) , (4,17)} and f (x) = 2x+1 , find the function 'g' in order  pair form.

    g = {(5,9) , (7,13) , (9,17)}


    g = {(5,5) , (7,3) , (2,17)}


    g = {(5,7) , (6,13) , (16,17)}


    g = {(5,8) , (7,14) , (0,17)}


  • If f = {(1,5) , (2,4) , (3,6)} and fog = {(5,6) , (4,5) , (6,4)} then find g (^-1) .

    g (^-1) = {(5,3) ,(4,4) (,6,2)}


    g (^-1) = {(3,5) ,(1,4) (2,6)}


    g (^-1) = {(1,2) ,(1,4) (2,6)}


    g (^-1) = {(2,5) ,(4,4) (4,6)}


  • If f = {(2,3) (4,5) , (5,6)} and g = {(4,3) , (2,5) , (5,8)} then find gof  (^-1) in ordered pair form .

    {(3,5) , (5,3) , (5,3)}


    {(1,5) , (2,5) , (,3,8)}


    {(5,5) , (3,3) , (8,8)}


    {(3,5) , (5,3) , (6,8)}


  • If f (x) = 2x -1 and g (x) = x+1 , find gof  (x) and gof  (-2) .

    4x, -2


    2x , -4


    x2 , -2


    3x, -3


  • If f (x) = 4x , g(x) = x+1 and fog(x) = 20 , find the value of x.

    4


    80


    8


    2


  • If a function f (x) = 5 then find f (4) and ff (-5) .

    -5


    -4,-5


    5,5


    4,5


  • If f (x) = 3x-1 and fg(x) = 9x-7 , then find g(x) .

    2x - 3


    x-2


    23 - x


    3x - 2


  • If f (x) = 5 - (frac{6}{x}) then what is the value of x at f (x) = -f (^-1) (-1) ?

    1


    (frac{-1}{-5})


    -1


    -5


  • If f (x) = (frac{6}{x-2}) , g(x) = k x (^2) -1  and gof (5) = 7 , what is the value of k ?

    30


    12


    2


    35


  • If f (x) = 4x+5 and fog (x) = 8x + 13 , find the value of x such that gof (x) = 28 .

    -2


    16


    1


    2


  • If f (x) = 2x + 3 and fog (x) = 2x + 5 , find (gof) (^-1) .

    (frac{4-x}{2})


    (frac{x-4}{2})


    (frac{x-2}{2})


    (frac{x-4}{4})


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Find the value of k if (k 2), (4k-6), (3k-6) are first term of an AP.


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Find the nth term of the sequence 20, 15, 10, 5, ...........


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