20 Class 10 Semester 2 Mathematics

7/31/2019 20 Class 10 Semester 2 Mathematics

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A quadratic equation is an equation of the form

Where a, b, c are given number (constant) and , and x is an unknownquantity(variable).

The number a, b and c are called the coefficient of the quadratic equation. a is the

coefficient of, b is the coefficient of x and c is called the constant term.Ex:- (i) is a quadratic equation. here a=3, b=2 and c=-1.

(ii) is not a quadratic equation.

An expression of the form , where a, b, c are constant and is called aquadratic expression.

What is the difference between a function and an equation? What we did in the part

above was quadratic equation. In an equation, we have something equal to zero. But in a

function we just have the expression. Now we look at raw function or expression

and not

When we are dealing with

,

we try to find the values of x where the given expression takes value 0.When we just look at the function or the expression we simply see itsvalue at different values of x. Like at x=1 or x=2 or x=0 or x=-b/2a or at x=

.For each of the above, the answers are a+b+c, 4a+2b+c, c-a, and 0respectively

Suppose a charity trust decides to build a prayer hall having a carpet area of

300 square metres with its length one metre more than twice its breadth. Whatshould be the length and breadth of the hall?

How we should proceed? We can assume that size of breadth=x meter. Then

length should be (2x+1)meter. So, its area is x(2x+1) i.e or is quadratic equation where a=2, b=1, c=-300.

:Is the equation a quadratic equation?


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The given equation is

We can also write it as

Hence, we can say that the given equation is a quadratic equation.

If two expressions in x are equal for all values of x, this statement is called an identity.

(i) (ii) Note:-

(i) An equation in x is true for some particular values of x where as an identity in

true for all values of x.

(ii) the symbol is used to distinguish an identity from an equation.


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The values of the x which satisfy the equation are called roots of theequation

. roots of equation

are also called of

the quadratic expression .The solution is not very tough if you dont get scared and dont fear reading toughlooking things.

What we are trying to do in the steps above is express the quadratic equation as the

square where k is a constant.

We are trying to solve ,which is same as trying to solve for { }

is called the of quadratic equation and isdenoted by the capital letter D.

We are pretty sure that some of you must not have read the solution above. If you are

one of those who understood or knew the proof before hand, then it is ok, but if you

have not been able to understand the proof and have ignored it for the fear of it, please

read it. The understanding and the motivation of the result is very very important. It

will be a very good tool for you in some more complex problem at some other time.


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Sum of roots

Product of roots

( ) ( )

If are the roots of equation , find the value offollowing expression

(i) (ii) (iii) (iv) (v) In such type of problems, try to represent the given expression in terms of

(sum of roots) and (product of roots)In the given problem: (i) (ii)

(iii) ()(iv) (v) Using (i) & (ii)


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Let be the quadratic equation whose roots are Therefore,

and

Now required quadratic equation is

or

the required quadratic equation is will be sum of roots )x +product of roots =0If

are the roots of equation

, form an equation

whose roots are: (i) (ii) (i)

The equation is:

is the required equation.(ii)

The equation is: ()

is the required equation.

Form an equation whose roots are squares of the sum and the difference of

the roots of the equation Let are the roots of given equation,

Now we have to make an equation whose roots are


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The required equation is

: The roots of 2x2-7x are and find the equation with roots and. We have from equation and

Sum of roots for the required equation is

[] ,

Product of roots= So the equation is

or

7

If be the roots of equation , (where a, b, c are real) then Now there are four possibilities

- when i.e will be imaginaryHence from (1), will be both imaginary.

- when i.e ,

Hence both roots

will be real and equal.

:- when i.e >0 will be real. Let =dThen from (1),

and Hence the roots will be real and different (distinct).when D i.e is a perfect square and a, b, c are rational.

is a rational number, let =k


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Then from (i) Therefore, both the roots will be rational.

Comment upon the nature of roots:

(i) (ii) (iii)

(i) Find Discriminant. (D)

hence the roots are real.

(ii)

and also a perfect square, hence the roots are rational.(iii)

and also a prefect square, hence the roots are rational.

Irrational roots of a quadratic eq. with rational coefficient always occur in conjugate

pair i.e if be one of the equation, where a, b, c are rationalthen its other root will be . (This can be proved with some work but we will notneed the proof of the same. Interested readers can visit targetiit.com to get the proofs)

Find all the roots of the equation: 7 if oneroot is As the coefficients are real, complex roots will occur in conjugate pairs. Soother root is

Let be the remaining roots.the four roots are

Hence the factors are: ( ) ( )


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Dividing 7 by or by inspection wecan find that the other factor of quadratic equation is

7 are roots of = hence roots are

What can you say about the roots of the following equations?

(i) (ii)

(i) Calculate D

so the roots are complex of a = 5/3 and real and equal if a = 5/3.(ii) Simplifying the given equation: Now using the identity:

We get: so the roots are real.Note: if then ie, the roots are equal.

We have more or less done the foundations of this chapter Arent y ou surprised thatthis is the foundation of this chapter! The fact is that we have to get into the depth of

this chapter and see graphs and the graphical approach to solving such problems. If you

fear graphs, there are no worries as we will deal with graphs here itself so that you

dont have too much to worry aboutNow we come to the shape of the quadratic function. Having understood the subtle

difference between an expression and an equation, let us try to now explore what else

can be done of the expression other than finding its roots.


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: Find the roots of the equation .

First, we rearrange the given equation according to the standard format.

x2x= 12x2x 12 = 0We need a pair of numbers whose product is 12 and whose sum is 1.These numbers are 3 and

4.

We can now rewrite the equation as

x2x 12 = 0x2 + 3x 4x 12 = 0x(x+ 3) 4(x+ 3) = 0

(x

4)(x+ 3) = 0

The solutions of the given equation can now be determined by equating the two factors

to 0.

(x 4) = 0 and (x+ 3) = 0x= 4 and x= 3Thus, the two roots of the quadratic equation are 4 and 3.

: Find the roots of the equation .

First, we rearrange the given equation according to the standard format.

6x2 + 5x 4 = 0Here, we need to find two such numbers whose product is 6 (4) = 24 and whosesum is 5.

The required numbers are 8 and 3.


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Now, [8 (3) = 24 and 8 + (3) = 5]6x2 + 5x 4 = 0

6x2

+ 8x3x 4 = 0 2x(3x+ 4) 1(3x+ 4) = 0 (2x 1)(3x+ 4) = 0 2x 1 = 0 and 3x+ 4 = 0

Thus, the roots of the given equation are .

: Find two positive consecutive odd integers whose product is 99.

Let the first integer be x.

The next odd integer will thus be x+ 2.

x(x+ 2) = 99x2 + 2x 99 = 0x2 + 11x 9x 99 = 0x(x+ 11) 9(x+ 11) = 0(x 9)(x+ 11) = 0x 9 = 0 and x+ 11 = 0x= 9 and x= 11

However, we know that the two integers are positive.

x= 9Thus, the two integers are 9 and (9 + 2) = 11.


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: The width of a rectangle is 16 feet less than 3 times its length. If the area of

the rectangle is 35 ft2, then what are the dimensions of the rectangle?

Here, the width of the rectangle has been expressed in terms of its length.

Hence, let us take the length of the rectangle to bex.

Thus, width of the rectangle = 3x 16The area of the rectangle is 35 ft2.

35 = x(3x16)35 = 3x2

16x

3x2 16x 35 = 03x2 21x+ 5x 35 = 03x(x 7) + 5(x 7) = 0(3x+ 5)(x 7) = 03x+ 5 = 0 and x

7 = 0

However, the length of a rectangle cannot be negative.

x= 7Thus, length of the rectangle is x= 7 ft and the width of the rectangle is 3x 16 = (3 7 16) ft = 5 ft

: Find the roots of the equation by the method of completion of

squares.

The given equation is .

By multiplying the equation by 5, we get .


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: Find the roots of the equation by the method of completion of

squares.

The given equation is .

: Find three consecutive even integers so that the product of the first two

numbers is three times the third number.

Let the three numbers be x, (x+ 2), and (x+ 4).

From the given condition, we have x(x+ 2) = 3(x+ 4).


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However, the three numbers are positive integers.

Thus, xcannot be equal to 3.Thus, the three consecutive even integers are 4, 6, and 8.

: Solve the quadratic equation 5x2 + x by using the quadratic formulaOn comparing the given equation with the general form ax2 + bx+ c= 0, we

get

a= 5, b= 1, c On putting these values in the quadratic formula, we get


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: Find the roots of the equation 3x2 + 5x+ 10 = 0 by using the quadratic

equation formula.

On comparing the given equation with the general formax2 + bx+ c= 0, we

get

a= 3, b= 5, c= 10

b2 ac= (5)2 span Thus, does not have any real value.

Thus, there are no real roots for the given equation.

: The denominator of a fraction is 2. If the sum of the fraction and its reciprocal

is , then find the fraction.

Let the numerator of the fraction be x.

Hence, the rational number is .


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It is given that the sum of the fraction and its reciprocal is .

On comparing this equation with the general form ax2 + bx+ c= 0, we get

a= 6, b c= 24On putting these values in the quadratic formula, we get

Since xis the numerator of the fraction, it has to be an integer.

x= 3Thus, the fraction is .

: The perimeter of a rectangle is 34 cm and its area is 70 cm2. Find the length

and the breadth of the rectangle.


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Let the length and breadth of the rectangle be xandyrespectively.

Thus, perimeter of the rectangle = 2(x+y) = 34 cm

x+y= 17y= 17 x Thus, area of the rectangle = xy= 70 cm2 Substituting the value ofyfrom (1) in (2):

x(17 x) = 70

17xx2 = 70x2 17x+ 70 = 0By using the quadratic formula, we get

Thus, the length and the breadth of the rectangle are 10 cm and 7 cm respectively.

: Both the roots of given equation 0))(())(())(( axcxcxbxbxax are

always

(a)Positive (b)Negative (c)Real (d) Imaginary

(c) Given equation 0))(())(())(( axcxcxbxbxax can be re-written as

0)()(232

cabcabxcbax

][4)](3)[(42222

acbcabcbacabcabcbaD

0])()()[(2222 accbba


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Hence both roots are always real.

: If the roots of 0)()()( 2 baxacxcb are equal then ca

(a)2b (b) 2b (c)3b (d) b

(a) 0 baaccb

Hence one root is 1. Also as roots are equal, other root will also be equal to 1.

Alsocb

ba

.

cb

ba

1.1 cbba cab 2

: If the roots of equationrqxpx

111

are equal in magnitude but opposite in

sign, then )( qp

(a)2r (b)r (c) 2r (d) None of these(a) Given equation can be written as 0])([)2(2 rqppqxrqpx

Since the roots are equal and of opposite sign,

Sum of roots = 0 0)2( rqp rqp 2

: If 3 is a root of 0242 kxx , it is also a root of

(a) 052 kxx (b) 052 kxx (c) 062 kxx (d) 0242 kxx

(c) Equation 0242 kxx has one root as 3,

024332 k 5k

Put 3x and 5k in option

Only (c) gives the correct answer i.e. 091532 00

: For what values of k will the equation 0)23(7)31(22 kxkx have equal roots

(a)1, 10/9 (b)2, 10/9 (c)3, 10/9 (d) 4, 10/9(b) Since roots are equal then )23(7.1.4)]31(2[ 2 kk kkk 1421691 2 02089 2 kk

Solving, we get 9/10,2 k

Solve the quadratic equation .

The given quadratic equation is .

The discriminant of this equation is

b2 ac=Thus, the solution of the given equation is


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If the roots of the quadratic equationax2 + bx+ c= 0 are imaginary, then

what can we say about the signs ofaand c?

The roots of quadratic equation ax2 + bx+ c= 0 are imaginary if the

discriminantb2 ac< 0.Here, b2 is always positive whatever the sign ofbis. Hence, the discriminant is negative

if the productacis positive. Thus, aand cmust have the same signs.

: Find the value of discriminant for the quadratic equationx2 + 4x The given quadratic equation is x2 + 4x 0.

On comparing this equation with the standard form of quadratic

equation, ax2 + bx+ c= 0, we have

a= 1, b= 4, c Therefore, the discriminant, b2 ac= (4)2 = 16 + 84

= 100

Thus, the value of discriminant for the given quadratic equation is 100.

: Find the nature of the roots of quadratic equation, x2 x+ 16 = 0.The given equation is x2 x+ 16 = 0.

On comparing this equation with the standard form of quadratic

equation, ax2 + bx+ c= 0, we have a= 1, b c= 16.Therefore, the discriminant,

b2 ac 2 b2 ac b2 ac= 0Thus, the given quadratic equation has two equal real roots.


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Find the value ofkfor which the quadratic equation x2 x+ k= 0 has tworeal equal roots.

On comparing the given equation with ax2 + bx+ c= 0, we have a= 1, b

12, c= kFor two real equal roots,

b2 ac= 0 2 k= 0 k= 0 4k= 144k= 36Thus, the value ofkis 36.

Is it possible to design a rectangular garden whose area is 250 square m and

the sum of whose length and breadth is 20 m?

Let the length of the garden be xm.Breadth x) mArea of garden = length breadth

250 = x x) 250 = 20xx2

x2 x+ 250 = 0This is a quadratic equation whose discriminant is b2 ac 2 Therefore, the roots of the equation are not real. Hence, it is not possible to design such

a garden.


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Can we find two consecutive positive numbers whose product is 420? If yes,

then find the numbers.

Let one number be x.

Then the other number = x+ 1

According to the question,

x (x+ 1) = 420

x2 + x= 420

x2 + x

The discriminant of this equation is b2 ac= (1)2 = 1 + 1680

= 1681

b2 ac> 0Therefore, the roots of the quadratic equationx2 + x

are real

We can find the numbers using quadratic formula as follows

x and We take x as is a negative number.Thus, 20 and 21 are the two consecutive positive numbers whose product is 420.

: An express train takes 1 hour less than a passenger train to travel a distanceof 132 km. If the average speed of the express train is 11 km/h more than that of a


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passenger train, then form a quadratic equation to find the average speed of the express

train?

Let the average speed of the express train be xkm/h.

Since it is given that the speed of the express train is 11 km/h more than that of a

passenger train,

Therefore, the speed of the passenger train will be x kmhAlso we know that

Time taken by the express train to cover 132 km =

Time taken by the passenger train to cover 132 km =

And the express train takes 1 hour less than the passenger train. Therefore,

This is the required quadratic equation.

: A chess board contains 64 equal squares and the area of each square is 6.25

cm2. A border 2 cm wide is made around the board. Formulate a quadratic equation to

find the length of each side of the board?

Let the length of each side of the chess board be xcm.

Therefore, the length of the chess board without border will be x cm as shown in thefollowing figure.


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Area of the chess board without border = (x 2It is given that the area of each square is 6.25 cm2.

Area of 64 squares on the chess board = 6.25 64

Therefore,

(x 2 = 64 6.25

This is the required quadratic equation.


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Complex Numbers

An Imaginary Number, when squared, gives a negative result.

Let's try squaring some numbers to see if we can get a negative result:

2 2 = 4

(-2) (-2) = 4 (because a negative times a negative gives a positive) 0 0 = 0 0.1 0.1 = 0.01

No luck! Always positive, or zero.

It seems like you cannot multiply a number by itself to get a negative answer ...

... but imagine that there is such a number (call it i for

imaginary) that could do this:

i i = -1

Would it be useful, and what could we do with it?

Well, by taking the square root of both sides we get this:

Which means that i is the answer to the square root of -1.

Which is actually very useful because ...

... by simply accepting that i exists lets us solve things

where we need the square root of a negative number.
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: What is the square root of -9 ?

= (9 -1) = (9) (-1) = 3 (-1) = 3i

Hey! that was interesting! The square root of -9 is simply the square root of +9, times i.

In general:

(-x) = ix

So long as we keep that little "i" there to remind us that we still

need to multiply by -1 we are safe to continue with our solution!

Using i we can also come up with new solutions:

: Solve x2 = -1

Using Real Numbers there is no solution, but now we can solve it:

Answer: x = -1 = i

Unit Imaginary Number

The "unit" Imaginary Number (the equivalent of 1 for Real Numbers)is (-1) (the square root of minus one).

In mathematics we use i (for imaginary) but in electronics they

use j (because "i" already means current, and the next letter after i is

j).

Examples of Imaginary Numbers

i 12.38i -i 3i/4 0.01i -i/2

Imaginary Numbers are not "Imaginary"

Imaginary Numbers were once thought to be impossible, and so they were called

"Imaginary" (to make fun of them).

But then people researched them more and discovered they were

actually useful and important because they filled a gap in mathematics ... but the

"imaginary" name has stuck.
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And that is also how the name "Real Numbers" came about (real is not imaginary).

Here are 2 cases where they are useful:

AC (Alternating Current) Electricity changes between positiveand negative in a sine wave.

If you combine two AC currents they may not match properly,

and it can be very hard to figure out the new current.

But using imaginary numbers and real numbers together

makes it a lot easier to do the calculations.

And the result may be "Imaginary" current, but it could still

hurt you!

Quadratic Equation

The Quadratic Equation can give results

that include imaginary numbers

Interesting Property

The Unit Imaginary Number, i, has an interesting property. It "cycles" through 4

different values each time you multiply:

i i = -1, then -1 i = -i, then -i i = 1, then 1 i = i (back to i again!)

So we have this:

i = -1 i2

= -1 i3

= --1 i4

= 1 i5

= -1 ...etc
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: What is i6 ?

i6 = i4 i2

= 1 -1

= -1

The unit imaginary number, i, equals the square root of minus 1

Imaginary Numbers are not "imaginary", they really exist, and you

may need to use them one day.

A Complex Number is a combination of:

Real Numbers are just numbers like:

1 12.38 -0.8625 3/4 2 1998

Nearly any number you can think of is a Real Number

Imaginary Numbers are special because:

When squared, they give a negative result.

Normally this doesn't happen, because:

when you square a positive number you get a positive result, and when you square a negative number you also get a positive result (because a

negative times a negative gives a positive)

But just imagine there is such a number, because we are going to need it!

The "unit" imaginary number (like 1 for Real Numbers) is i, which is the square root of -

1
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(Read Imaginary Numbers to find out more.)

So we have this definition:

A Complex Number is a combination of a Real Number and an Imaginary Number

:

1 + i 39 + 3i 0.8 - 2.2i -2 + i 2 + i/2

Can a Number be a Combination of Two Numbers?

Can you make up a number from two other numbers? Sure you can!

You do it with fractions all the time. The fraction 3/8 is a number

made up of a 3 and an 8. We know it means "3 of 8 equal parts".

Well, a Complex Number is just two numbers added together (a Real and an Imaginary

Number).

Either Part Can Be Zero

So, a Complex Number has a real part and an imaginary part.

But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex

Numbers.
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Complex

NumberReal Part Imaginary Part

3 + 2i 3 2

5 5 0

-6i 0 -6

Complex does not mean complicated.

It means the two types of numbers, real and imaginary, together form a complex, just

like you might have a building complex (buildings joined together).

To add two complex numbers we add each element separately:

(a+bi) + (c+di) = (a+c) + (b+d)i

: (3 + 2i) + (1 + 7i) = (4 + 9i)

To multiply complex numbers:

Each part of the first complex number gets multiplied by

each part of the second complex number

Just use "FOIL", which stands for "Firsts, Outers, Inners, Lasts" (see Binomial

Multiplication for more details):

Firsts: ac Outers: adi Inners: bic Lasts: bidi

(a+bi)(c+di) = ac + adi + bci + bdi2

Like this:
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: (3 + 2i)(1 + 7i)

(3 + 2i)(1 + 7i) = 31 + 37i + 2i1+ 2i7i

= 3 + 21i + 2i + 14i2

= 3 + 21i + 2i - 14 (because i2

= -1)= -11 + 23i

Here is another example:

: (1 + i)2

(1 + i)2 = (1 + i)(1 + i) = 11 + 1i + 1i + i2

= 1 + 2i - 1 (because i2 = -1)

= 0 + 2i

But There is a Quicker Way!

Use this rule:

(a+bi)(c+di) = (ac-bd) + (ad+bc)i

: (3 + 2i)(1 + 7i) = (31 - 27) + (37 + 21)i = -11 + 23i

Why Does That Rule Work?

It is just the "FOIL" method after a little work:

(a+bi)(c+di) = ac + adi + bci + bdi2 FOIL method

= ac + adi + bci - bd (because i2=-1)

= (ac - bd) + (ad + bc)i (gathering like terms)

And there you have the (ac - bd) + (ad + bc)i pattern.

This rule is certainly faster, but if you forget it, just remember the FOIL method.

Let us try i2

Just for fun, let's use the method to calculate i2

: i2

i can also be written with a real and imaginary part as 0 + i

i2 = (0 + i)2 = (0 + i)(0 + i)

= (00 - 11) + (01 + 10)i

= -1 + 0i= -1


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And that agrees nicely with the definition that i2 = -1

So it all makes sense!

A conjugate is where you change the sign in the middle like this:

A conjugate is often written with a bar over it:

:

5 - 3i = 5 + 3i

The conjugate is used to help division.

The trick is to multiply both top and bottom by the conjugate of the bottom.

: Do this Division:

2 + 3i

4 - 5i

Multiply top and bottom by the conjugate of 4 - 5i :

2 + 3i

4 + 5i=

8 + 10i + 12i + 15i2

4 - 5i 4 + 5i 16 + 20i - 20i - 25i2

Now remember that i2 = -1, so:

=

8 + 10i + 12i - 15

16 + 20i - 20i + 25

Add Like Terms (and notice how on the bottom 20i - 20i cancels out!):

=-7 + 22i

41

We should then put the answer back into a + bi form:

=-7

+22

i41 41

DONE!

Yes, there is a bit of calculation to do. But it can be done.
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You can save yourself a little bit of time, though.

In that example, what happened on the bottom was interesting:

(4 - 5i)(4 + 5i) = 16 + 20i - 20i - 25i2

The middle terms cancel out!

And since i2=-1 we ended up with this:

(4 - 5i)(4 + 5i) = 42 + 52

Which is really quite a simple result

In fact we can write a general rule like this:

(a + bi)(a - bi) = a2 + b2

Remember that when you do division ... it will save you time.

So we could have done it like this:

: Do this Division:

2 + 3i

4 - 5i

Multiply top and bottom by the conjugate of 4 - 5i :

2 + 3i

4 + 5i=

8 + 10i + 12i + 15i2=

-7 + 22i

4 - 5i 4 + 5i 16 + 25 41

And then back into a + bi form:

=-7

+22

i41 41

DONE!


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The modulus of a complex number z = a + ib is denoted by and defined as.

o For example: The modulus of the complex numberz = is

The following results hold true for two complex numbersz1 andz2.o

o

The conjugate of a complex numberz = a + ib is denoted by and defined as.

o For example: The conjugate of the complex number is

The following results hold true for two complex numbersz1 andz2.ooo

o The modulus of a complex number and the modulus of its conjugate are equal.

The multiplicative inverse of a complex numberz = a + ib is given by

is the conjugate and is the modulus of the complex


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numberz.

Or

This is the required relation.

The beautiful Mandelbrot Set (pictured

here) is based on Complex Numbers.

It is a plot of what happens when you take

the simple equation z2+c (both complex

numbers) and feed the result back

into z time and time again.

The color shows how fast z2+c grows, and

black means it stays within a certain range.

Here is an image made by zooming into the

Mandelbrot set


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And here is the center of the previous onezoomed in even further:

Which of the following numbers is not a complex number?

can be written as , which is of the form a+ ib. Thus, is a

complex number.

is not of the form a+ ib. Thus, is not a complex number.

1 5iis of the form a+ ib. Thus, 1 5iis a complex number.

What are the real and imaginary parts of the complex number ?

The complex number can be written as , which is of

the form a+ ib.

Re z= a= and Im z= b=

For what values ofxandy, z1 = (x+ 1) 10iand z2 = 19 + i(yx)

represent equal complex numbers?

Two complex numbers are equal if their corresponding real and imaginary

parts are equal.

For the given complex numbers,

x+ 1 = 19 andyx= 10


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x= 18 andy 18 = 10

x= 18 andy= 8

Thus, the values ofxandyare 18 and 8 respectively.

: Perform the indicated operation and write the answers in standard form.

(a) (4+) +(510)

(b) (4+12) (315)

(c) 5 ( 9 + )There really isnt much to do here other than add or subtract. Note that the

parentheses on the first terms are only there to indicate that were thinking of that term

as a complex number and in general arent used.

(a)

(b)

(c)

: IfZ1= 3 i andZ2 = 1 + 2i, then write the complex number (Z1 + 2Z2 4) in the

form a + ib and determine the values ofa and b.

Solution: We haveZ1= 3 i andZ2 = 1 + 2i

Z1 + 2Z2 4 = (3 i) + 2 (1 + 2i) 4

= 3 i + 2 + 4i 4

= 1 + 3i

Which is of the form a + ib

a = 1 and b = 3

: What is the additive inverse of ?

Solution: Let

Additive inverse of


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: Multiply each of the following and write the answers in standard form.

(a)

(b)

(c)

(d)

(a) So all that we need to do is distribute the 7i through the parenthesis.

Now, this is where the small difference mentioned earlier comes into play. This number

is NOT in standard form. The standard form for complex numbers does not have an i2 in

it. This however is not a problem provided we recall that

Using this we get,

We also rearranged the order so that the real part is listed first.

(b) In this case we will FOIL the two numbers and well need to also remember to get

rid of the i2.

(c) Same thing with this one.

(d) Heres one final multiplication that will lead us into the next topic.

Dont get excited about it when the product of two complex numbers is a real

number. That can and will happen on occasion.

Simplify the following:


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Ifx+ iy= (2 + 5i) (7 + i), then what are the values ofxandy?

On equating the real and imaginary parts, we obtain

x= 9 andy= 37

What is the value of ?

We know that


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Students may make mistakes while solving this question.

We know that . However, when aand bare both negative,

then .

Hence, this question cannot be solved as

What is the multiplicative inverse of 5 9i?

Letz= a+ ib= 5 9i

Accordingly, a= 5 and b= 9

We know that


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Thus, is the multiplicative inverse of 5 9i.

: Write each of the following in standard form.

(a)

(b)

(c)

(d)

So, in each case we are really looking at the division of two complex

numbers. The main idea here however is that we want to write them in standard

form. Standard form does not allow for any i's to be in the denominator. So, we need to

get the i's out of the denominator.

This is actually fairly simple if we recall that a complex number times its conjugate is a

real number. So, if we multiply the numerator and denominator by the conjugate of the

denominator we will be able to eliminate the i from the denominator.

Now that weve figured out how to do these lets go ahead and work the problems.

(a)

Notice that to officially put the answer in standard form we broke up the fraction into

the real and imaginary parts.

(b)


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(c)

(d) This one is a little different from the previous ones since the denominator is a pure

imaginary number. It can be done in the same manner as the previous ones, but there is

a slightly easier way to do the problem.

First, break up the fraction as follows.

Now, we want the i out of the denominator and since there is only an i in the

denominator of the first term we will simply multiply the numerator and denominator

of the first term by an i.

: Write the complex number in the form of a + ib.

Solution:


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: Multiply the following and write the answer in standard form.

If we where to multiply this out in its present form we would get,

Now, if we were not being careful we would probably combine the two roots in the finalterm into one which cant be done!

So, there is a general rule of thumb in dealing with square roots of negative

numbers. When faced with them the first thing that you should always do is convert

them to complex number. If we follow this rule we will always get the correct answer.

So, lets work this problem the way it should be worked.

:Determine the conjugate and multiplicative inverse of .

Solution: Let

Accordingly, conjugate, and

Now, the multiplicative inverse is given by

: What is the conjugate of ?

Solution: Letz =

In order to find the conjugate ofz, we first write it in the form ofa + ib.


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Now,

Thus, the conjugate of the given complex number is .

: What is the modulus of ?

Solution: Modulus,

It can be written as

Continuing in this manner, we can write

Now,


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: Solve the quadratic equation .

Solution: The given quadratic equation is .

The discriminant of this equation is

b2 4ac =

Thus, the solution of the given equation is

: If the roots of the quadratic equation ax2 + bx + c = 0 are imaginary, then what

can we say about the signs ofa and c?

Solution: The roots of quadratic equation ax2 + bx + c = 0 are imaginary if the

discriminant b2 4ac < 0.

Here, b2 is always positive whatever the sign ofb is. Hence, the discriminant is negative if the

product ac is positive. Thus, a and c must have the same signs.


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A.P and G.P

In an Arithmetic Sequence the difference between one term and the next is a constant.

In other words, you just add the same value each time ... to infinity.

:

1, 4, 7, 10, 13, 16, 19, 22, 25, ...This sequence has a difference of 3 between each number.

In General you could write an arithmetic sequence like this:

{a, a+d, a+2d, a+3d, ... }

where:

a is the first term, and

d is the difference between the terms (called the "common difference")

: (continued)

1, 4, 7, 10, 13, 16, 19, 22, 25, ...

Has:

a = 1 (the first term)

d = 3 (the "common difference" between terms)

And we get:

{a, a+d, a+2d, a+3d, ... }

{1, 1+3, 1+23, 1+33, ... }

{1, 4, 7, 10, ... }


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You can write an Arithmetic Sequence as a rule:

xn = a + d(n-1)

(We use "n-1" because d is not used in the 1st term).

: Write the Rule, and calculate the 4th term for

3, 8, 13, 18, 23, 28, 33, 38, ...

This sequence has a difference of 5 between each number.

The values of a and d are:

a = 3 (the first term) d = 5 (the "common difference")

The Rule can be calculated:

xn = a + d(n-1)

= 3 + 5(n-1)

= 3 + 5n - 5

= 5n - 2

So, the 4th term is:

x4 = 54 - 2 = 18

Is that right? Check for yourself!Arithmetic Sequences are sometimes called Arithmetic Progressions (A.P.s)

To sum up the terms of this arithmetic sequence:

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

use this formula:

What is that funny symbol? It is called Sigma Notation

(called Sigma) means "sum up"

And below and above it are shown the starting and ending values:
http://www.mathsisfun.com/algebra/sigma-notation.htmlhttp://www.mathsisfun.com/algebra/sigma-notation.html


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It says "Sum up n where n goes from 1 to 4. Answer=10

Here is how to use it:

: Add up the first 10 terms of the arithmetic sequence:

{ 1, 4, 7, 10, 13, ... }

The values of a, d and n are:

a = 1 (the first term) d = 3 (the "common difference" between terms) n = 10 (how many terms to add up)

So:

Becomes:

= 5(2+93) = 5(29) = 145

Check: why don't you add up the terms yourself, and see if it comes to 145

: Check whether the following sequences form an A.P. or not.

(i)

(ii)

(iii)

(iv) A list of prime numbers greater than 2

(v) A list of the squares of natural numbers


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(i)The given sequence is

Difference between the second and the first term

Difference between the third and the second term

Difference between the fourth and the third term

Since the difference between any two consecutive terms is a constant, the sequence is

an arithmetic progression.

(ii)The given sequence is




Since the difference between any two consecutive terms is a constant, the sequence is

an arithmetic progression.

(iii)The given sequence is

This sequence can be rewritten as


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Since the difference between the consecutive terms is not a constant, the sequence

is not an arithmetic progression.

(iv)The list of the prime numbers greater than 2 is: 3, 5, 7, 11

Difference between the second and the first term = 5 3 = 2

Difference between the third and the second term = 7 5 = 2

Difference between the fourth and the third term = 11 7 = 4



(v)The list of the squares of natural numbers is: 12

, 22

, 32

, 42

= 1, 4, 9, 16

Difference between the second and the first term = 4 1 = 3

Difference between the third and the second term = 9 4 = 5

Difference between the fourth and the third term = 16 9 = 7



: Find the common difference of the A.P.

Also, state whether it is finite or infinite and write

down its first term.

The given A.P. is

a1 =

a2 =


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a3 =

a4 =

Common difference of the A.P. =

First term of the A.P. =

The given A.P. is infinite because its last term cannot be calculated.

: State whether the following statements are true or false.

(i) The sequence forms an A.P. with the first

term as and the common difference as .

(ii) The common difference of the A.P. is .

(iii) The nth term of an A.P. is given by 3n 4. Its common difference is 4.

(i)The given sequence is

Difference between the first and the second term

Difference between the second and the third term

Difference between the third and the fourth term

Since the difference between the consecutive terms of the sequence is constant, the

given sequence is an A.P. with the first term as and the common difference as

.

Thus, the statement is false.


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(ii)The given A.P. is

Common difference of the A.P. =

Thus, the statement is true.

(iii)It is given that the nth term of the A.P. is given by 3n 4.

Thus, first term of the A.P. = a = 3 1 4 = 3 4 = 1

Second term of the A.P. = a2 = 3 2 4 = 6 4 = 2

Third term of the A.P. = a3 = 3 3 4 = 9 4 = 5

Fourth term of the A.P. = a4 = 3 4 4 = 12 4 = 8

Thus, common difference of the A.P. = 8 5 = 5 2 = 2 (1) = 3

Thus, the statement is false.

: Find the missing terms in the following arithmetic progressions.

(i)

(ii)

(i) In the given A.P.,


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Thus, missing term, a3

(ii) In the given A.P.,

Thus, missing term, a2 =

: Find the first two terms of an A.P. in which the third term is 10 and the

common difference is 7.

Here, d= 7 and a3 = 10

Hence, the value ofa2 can be found by subtracting 7 from a3.

a2= 10 7 = 3

Similarly, the value ofa1 can be found by subtracting 7 from a2.

a1= 3 7 = 4

Thus, the first two terms of the A.P. are 4 and 3.

: Find the 20th term of the following arithmetic progression.

0.4, 1.5, 2.6, 3.7, 4.8

Here, a= 0.4 and d= 1.5 0.4 = 1.1


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Thus, the 20th term is given by,

a20 = a+ (20 1) d

= 0.4 + (20 1) 1.1

= 0.4 + 19 1.1

= 0.4 + 20.9

= 21.3

Thus, the 20th term of the given A.P. is 21.3.

: If the 7th term of an A.P. is 21 and 15th term is 53, then find the first termand common difference.

Let the first term and common difference of the A.P. be aand drespectively.

It is given thata7 = 21 and a15 = 53

Using the formula for nth term, we obtain

a7 = a+ (7 1) d

21 = a+ 6d (1)

And, a15 = a+ (15 1) d

53 = a+ 14d (2)

On subtracting equation (1) from (2), we obtain

32 = 8d

d= 4

Putting the value ofdin equation (1), we obtain

21 = a+ 6 (4)

21 = a 24

a= 3


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Thus, the first term is 3 and the common difference is 4.

: Is 102 a term of the A.P., 5, 11, 17, 23 ?

Let 102 be the nth

term of the given sequence.

an= 102

Using the formula for nth term, we obtain

an= a+ (n 1) d

102 = a+ (n 1) d

For the given A.P., a= 5 and d= 11 5 = 6

102 = 5 + (n 1) 6

102 5 = (n 1) 6

97 = (n 1) 6

However, nshould be a positive integer. Therefore, 102 is not a term of the given A.P.

: Find the number of three-digit numbers that are divisible by 5.

The first three-digit number which is divisible by 5 is 100, second is 105, third

is 110, and so on. The last three-digit number which is divisible by 5 is 995.

Thus, we obtain the following A.P.

100, 105 995

Here, we have to find the number of terms, n.

Last term of A.P. = 995


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The number of terms in the A.P. is n, so the last term is the nth term.

a+ (n 1)d= 995

Here,a

= 100 andd

= 5

100 + (n 1)5 = 995

(n 1)5 = 995 100

5n 5 = 895

5n= 895 + 5

5n= 900

n= 180

Thus, there are 180 three-digit numbers, which are divisible by 5.

: The fare of a bus is Rs 10 for the first kilometre and Rs 5 for each additional

kilometre. Find the fair after 12 kilometres.

The fare after each kilometre forms an A.P. as follows.

Fare after one kilometre = Rs 10

Fare after two kilometres = 10 + 5 = Rs 15

Fare after three kilometres = 15 + 5 = Rs 20

Now the arithmetic progression is 10, 15, 20

Here, first term, a= 10 and common difference, d= 5

Now the fare after 12 kilometres is the 12th term of the A.P.

a12 = a+ (12 1) d

a12 = 10 + 11 5

= 10 + 55


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= 65

Thus, the fare after 12 kilometres is Rs 65.

: Mohit borrowed a sum of money at a simple interest rate of 2% per annum.He has to pay an amount of Rs 1120 after 6 years. How much money did he borrow?

Let the amount of money Mohit borrowed be Rs x. We know that the amount

after Tyears is

Where, Pand Rdenotes the principal and rate respectively

The amount after every year forms an A.P.

Amount after first year

Amount after second year

Thus, the A.P. is as follows.

Here, the first term is and common difference is .

Now, it is given that the amount after 6 years is Rs 1120 i.e., 6th term of the A.P. is 1120.

Now using the formula, an= a+ (n 1)d, we obtain


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1120 100 = 112x

x= 1000

Thus, Mohit borrowed Rs 1000.

: Find the sum of first 25 terms of the following A.P.

2, 7, 12

Here, a = 2 and d = 7 2 = 5.

Sum of the first 25 terms is given by

= 25 62

= 1550

Thus, the sum of first 25 terms of the given A.P. is 1550.

: Find the sum of first 8 terms of the A.P whosenthterm is given by 6n 5.

The nth term is given by

an = 6n 5


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On replacing n by 1, 2, 3 respectively, we get the first, second, third terms of the A.P.

a1= 6(1) 5 = 1

a2= 6(2) 5 = 7

a3= 6(3) 5 = 13 and so on.

The A.P. so obtained is as follows.

1, 7, 13

Here, the first term, a = 1

and the common difference d = a2 a1 = 7 1 = 6

Using the formula, , the sum of first 8 terms is given by

= 176

Thus, the sum of first 8 terms is 176.

: How many terms of the A.P. 28, 24, 20 should be taken so that the

sum will be zero?

Let the sum of n terms be zero.

Here, a = 28 and d = 24 (28) = 4

Sum of n terms of an A.P. is given by

But, it is given that the sum of n terms of the given A.P. is zero.


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Thus, the sum of 15 terms of the A.P is zero.

: Sapnas father planted 4 trees in his garden, when he was 22 years old. After

that, every year he planted one more tree than the number of trees he planted in the

previous year. How many trees will be there in his garden when he will become 40

years old?

We can write the given information in the form of an A.P. as follows

Number of trees he planted in the first year = 4

Number of trees he planted in the second year = 4 + 1 = 5

Number of trees he planted in the third year = 5 + 1 = 6

And so on.

Now, the A.P. is 4, 5, 6

He planted trees from the age of 22 years to 40 years, i.e. for 18 years.

Thus, we have to find the sum of 18 terms of this A.P.

Here, a = 4, d = 1

And, n = 18

Using the formula, , the sum of 18 terms of this A.P. is given by


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But we know that the sum of n terms of an A.P. is

Thus, the sum of (p + q) terms is

Thus, the sum of (p + q) terms is .

: Find the nth term of the A.P., the sum of whose n terms is n2 + 2n.

Let Sn be the sum of n terms.

It is given that the sum of n terms of the A.P. is n2 + 2n.

Sn = n2 + 2n (1)

On replacing n by (n 1) in the equation, we obtain

Let an be the nth term of the A.P.

Therefore, we can write

Sn = Sn1 + an

Thus, an = Sn Sn1

= n2 + 2n [(n 1)2 + 2(n 1)]

= n2 + 2n [n2+ 1 2n + 2n 2]

= n2 + 2n n2 + 1


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= 2n + 1

Thus, the nth term of the A.P is (2n + 1).

In a Geometric Sequence each term is found by multiplying the previous term by

a constant.

:

2, 4, 8, 16, 32, 64, 128, 256, ...

This sequence has a factor of 2 between each number.

Each term (except the first term) is found by multiplying the previous term by 2.

In General you could write a Geometric Sequence like this:

{a, ar, ar2, ar3, ... }

where:

a is the first term, and r is the factor between the terms (called the "common ratio")

: {1,2,4,8,...}

The sequence starts at 1 and doubles each time, so

a=1 (the first term)

r=2 (the "common ratio" between terms is a doubling)So we would get:

{a, ar, ar2, ar3, ... }

= {1, 12, 122, 123, ... }

= {1, 2, 4, 8, ... }

But be careful, r should not be 0:

When r=0, you get the sequence {a,0,0,...} which is not geometric


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You can also calculate any term using the Rule:

xn = ar(n-1)

(We use "n-1" because ar0 is for the 1st term)

:

10, 30, 90, 270, 810, 2430, ...


The values of a and r are:

a = 10 (the first term) r = 3 (the "common ratio")

The Rule for any term is:

xn = 10 3(n-1)

So, the 4th term would be:

x4 = 103(4-1) = 1033 = 1027 = 270

And the 10th term would be:

x10 = 103(10-1) = 1039 = 1019683 = 196830

A Geometric Sequence can also have smaller and smaller values:

Example:

4, 2, 1, 0.5, 0.25, ...

This sequence has a factor of 0.5 (a half) between each number.

Its Rule is xn = 4 (0.5)n-1

Because it is like increasing the dimensions in geometry:

a line is 1-dimensional and has a length of r

in 2 dimensions a square has an area of r2


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in 3 dimensions a cube has volume r3

etc (yes you can have 4 and more dimensions in mathematics).

Geometric Sequences are sometimes called Geometric Progressions (G.P.s)

A G.P. is written in its standard form as a, ar, ar2, ar3, ar4

o Here, is called the of the G.P.o Here, is called the of the G.P.

There are two types of geometric progressions: finite and infinite.o A has finite number of terms. In general, a

finite G.P. with nterms can be written as a, ar, ar2, ar3, ar4arn1.

o An has infinite number of terms. In general,an infinite G.P. can be written as a, ar, ar2, ar3, ar4arn1

There are two types of geometric series: finite and infinite.o The series a+ ar+ ar2 + ar3 + ar4+ +arn1 is a .o The series a+ ar+ ar2 + ar3 + ar4+ +arn1+ is an

.

When you need to sum a Geometric Sequence, there is a handy formula.

To sum:

a + ar + ar2 + ... + ar(n-1)

Each term is ark, where k starts at 0 and goes up to n-1

Use this formula:

a is the first term

r is the "common ratio" between terms

n is the number of terms

What is that funny symbol? It is called Sigma Notation
http://www.mathsisfun.com/algebra/sigma-notation.htmlhttp://www.mathsisfun.com/algebra/sigma-notation.html


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(called Sigma) means "sum up"

And below and above it are shown the starting and ending values:

It says "Sum up n where n goes from 1 to 4. Answer=10

The formula is easy to use ... just "plug in" the values of a, r and n

Example: Sum the first 4 terms of

10, 30, 90, 270, 810, 2430, ...


The values of a, r and n are:

a = 10 (the first term) r = 3 (the "common ratio") n = 4 (we want to sum the first 4 terms)

So:

Becomes:

You could check it yourself:

10 + 30 + 90 + 270 = 400

And, yes, it was easier to just add them in this case, because there were only 4 terms. But

imagine you had to sum up lots of terms, then the formula is better to use.


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Let's see the formula in action:

: Grains of Rice on a Chess Board

n our page Binary Digits we give an example of grains of rice on a

hess board. The question is asked:

hen you place rice on the chess board:

grain on the first square,

grains on the second square,

grains on the third and so on,

.

... doubling the grains of rice on each square ...

... how many grains of rice in total?

So we have:

a = 1 (the first term) r = 2 (doubles each time) n = 64 (64 squares on a chess board)

So:

Becomes:

= (1-264) / (-1) = 264 - 1

= 18,446,744,073,709,551,615

Which was exactly the result we got on the Binary Digits page (thank goodness!)

And another example, this time with r less than 1:

: Add up the first 10 terms of the Geometric Sequence that halves each time:

{ 1/2, 1/4, 1/8, 1/16, ... }

The values of a, r and n are:

a = (the first term) r = (halves each time) n = 64 (10 terms to add)
http://www.mathsisfun.com/binary-digits.htmlhttp://www.mathsisfun.com/binary-digits.htmlhttp://www.mathsisfun.com/binary-digits.htmlhttp://www.mathsisfun.com/binary-digits.html


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So:

Becomes:

Very close to 1.

(Question: if we continue to increase n, what would happen?)

Which of the following sequences is not a geometric progression?

i. 11, 2.2, 0.44, 0.088, 0.0176 ii. 17, 17.5, 18, 18.5, 19

iii. 81,iv.

i. The series 11, 2.2, 0.44, 0.088, 0.0176 is a G.P. since

ii. The series 17, 17.5, 18, 18.5, 19 is not a G.P. since

iii. The series 81, is not a G.P. since

and


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iv. The series 16, 4, 1, is a G.P. since

Write the first term, last term and the common ratio of the following

geometric progressions:

i.ii.

i. In the G.P. ,First term, a= 3

Last term, l= 1728

Common Ratio, r=

ii. In the G.P., ,First term, a=

Last term, l=

Common Ratio, r=

: A scientist kept a solution on fire. The original temperature of the solution

was 24C. He noticed that the temperature of the solution increased by 20% of the

original temperature every hour. Find the temperature of the solution after the 5th hour.


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Solution: It is given that the original temperature of the solution was 24C. It is also

given that the temperature of the solution increased by 20% of the original temperature

every hour. Hence, the temperature after the 1 st hour will be

Temperature after 2nd hour:

Temperature after 3rd hour:

And so on

Hence, the temperature of the solution noted after every hour will form a geometric

progression with

First term, a = 24

Common ratio, r= = 1.2

We know that the nth term of a G.P. is given by an

= arn1.

Also, the temperature of the solution after the 5th hour will be the 6th term of the G.P., which is

given by

a6 = 24 (1.2)

5

= 59.7 (approx.)

Thus, the temperature of the solution after the 5th hour will be 59.7C approximately.

: Find the sum of the sequence 0.4, 0.44, 0.444. up ton terms.

Solution: The given sequence is not a G.P. However, we can relate it to a G.P. by

writing the terms as

Sn= 0.4 + 0.44 + 0.444 + ton terms


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: Find the number of terms of the G.P. that are required for

giving the sum as .

Solution: Let n be the required number of terms.

It is given that

First term, a= 6

Common ratio, r=

We know that the sum ofn terms of a G.P. is given by Sn

= . Therefore,


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Thus, the required number of terms of the G.P. is 5.

: In a G.P., the ratio of the 11 th term and the 14th term is given by 27:125. Find the

18th term of the G.P. if the 5th term of the G.P. is .

Solution: Let the first term and the common ratio of the G.P. be a and rrespectively.


= arn1. Therefore,

a11 = ar10

a14 = ar13

It is given that the ratio of the 11th term to the 14th term is 27:125. Hence,


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It is also given that the 5 th term of the G.P. is . Hence,

Thus, the 18th term of the G.P. is given by

: A scientist kept a solution on fire. The original temperature of the solution was

24C. He noticed that the temperature of the solution increased by 20% of the original

temperature every hour. Find the temperature of the solution after the 5th hour.

Solution: It is given that the original temperature of the solution was 24C.

It is also given that the temperature of the solution increased by 20% of the original

temperature every hour. Hence, the temperature after the 1 st hour will be

Temperature after 2nd hour:

Temperature after 3rd hour:

And so on

Hence, the temperature of the solution noted after every hour will form a geometric

progression with


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First term, a = 24

Common ratio, r= = 1.2


= arn1.

Also, the temperature of the solution after the 5th hour will be the 6th term of the G.P., which is

given by

a6 = 24 (1.2)5 = 59.7 (approx.)

Thus, the temperature of the solution after the 5th hour will be 59.7C approximately.

: Find the sum of the sequence 0.4, 0.44, 0.444. up ton terms.

Solution: The given sequence is not a G.P. However, we can relate it to a G.P. by writing

the terms as

Sn= 0.4 + 0.44 + 0.444 + ton terms

: Find the number of terms of the G.P. that are required for

giving the sum as .

Solution: Let n be the required number of terms.

It is given that


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First term, a= 6

Common ratio, r=

We know that the sum ofn terms of a G.P. is given by Sn

= . Therefore,

Thus, the required number of terms of the G.P. is 5.

: In a G.P., the ratio of the 11th term and the 14th term is given by 27:125. Find the

18th term of the G.P. if the 5th term of the G.P. is .

Solution: Let the first term and the common ratio of the G.P. be a and rrespectively.


= arn1. Therefore,

a11 = ar10

a14 = ar13

It is given that the ratio of the 11th term to the 14th term is 27:125. Hence,


75/75

It is also given that the 5 th term of the G.P. is . Hence,

Thus, the 18th term of the G.P. is given by

20 Class 10 Semester 2 Mathematics

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Transcript of 20 Class 10 Semester 2 Mathematics