Category: Maths Notes PPT

Algebraic Of Complex Numbers

Have you ever heard of complex numbers? Do you know what an iota is? What kind of number is [sqrt{-2}]?  Does the number even exist? To get all your answers, let’s first understand the entire number system.

Number System

The number system is broadly divided into two parts: Real Numbers and Complex Numbers. 

1. Real numbers 

Real numbers are those which can be shown on a number line. On the other hand, complex numbers are those which can not be expressed on a number line or be experienced in real life. Real Numbers are further divided into two categories called rational and irrational numbers. 

Rational numbers are numbers which can be expressed as fractions and their denominators are not equal to 0. All the real numbers which are not rational are Irrational numbers. Rational numbers are made by dividing two integers. Integers include all negative and positive natural numbers along with zero. Integers are further divided into two sub-categories: whole numbers and natural numbers. Whole numbers are positive counting numbers along with 0. When you remove 0 from whole numbers, we obtain positive counting numbers which are known as natural numbers.

2. Complex Numbers

Complex numbers are also known as Imaginary numbers. Now that we know the definition of complex numbers and that complex numbers are the part of the Number System, let’s see some examples. 

All the negative numbers under root are imaginary numbers. 

[sqrt{-2}] and [sqrt{-2}] are two very different things. The first one is a real number. Since it’s a negative number under root the second one is a complex number. A complex number is represented in the following way: a+bi, where a is the real part and b is the imaginary part.

You can write the complex number [sqrt{-2}]  in  a+bi form. 0+2i is equal to [sqrt{-2}]. You must be wondering why are we using the symbol ’i’? What does it mean? Well, it is iota. Have you ever heard of iota? If not, then is all that you need to know about iota.

IOTA

Iota is a greek letter which is used to represent the imaginary part of a complex number. Iota(i) is considered to be the square root of -1. It may also be defined as a number whose square is -1.

i=[sqrt{-1}]

i²=-1

i³=i

i⁴=1

Algebraic Operations On Complex Numbers:

Four types of algebraic operations can be done on complex numbers. These four algebra of complex numbers are:

• Addition

• Subtraction

• Multiplication

• Division

There are several properties that algebra on imaginary numbers follow:

Closure law

The sum or product of two imaginary numbers will always get you an imaginary number.

Commutative Law

If you change the order of imaginary number while adding or multiplying the result will not change that is the answer you get will always be the same. 

Associative Law

If you add or multiply any three complex numbers in any order the result will always remain the same. 

Existence of Additive Identity

This property tells us that if we add zero to any complex we will get the same complex number. This shows that there’s a number that can be added to get the same number back. It is also known as zero complex number and is denoted as 0 (or 0 + i0).

Existence of Additive Inverse

A complex number has the opposite sign for its both real and imaginary parts. This is known as the Existence of  Additive inverse.

Multiplicative Identity

Multiplicative Identity is a property which talks about the existence of a complex number that when multiplied to another will get the same result. it is denoted as 1 (or 1 + i0)

Multiplicative Inverse

It is a property of any non- zero complex number to have a reciprocal. This is known as the multiplicative inverse.

Distributive Property

When you split the multiplication of a complex number by another term this property is known as the distributive property.

Note: Subtraction follows all the properties followed by addition.

Fun Facts:

1. Both real and imaginary parts are present in the square root of i.

2. The N-th root can have N number of unique solutions and any root of i has multiple unique solutions

3. The result may vary depending on whether i is present in the numerator or denominator in an imaginary fraction. 

4. When you raise i to the i power, the number you get is a real number.

5.  Numbers like [pi ], i and e are all related to one another.

Triangle is the smallest polygon which has three sides and three interior angles, consisting of 3 edges and 3 vertices. A triangle with vertices A, B and C is denoted as ∆ABC. In a triangle, 3 sides and 3 angles are referred to as the elements of the triangle. Angle sum property and exterior angle property are the two important attributes of a triangle. 

In this article, we are going to learn the interior angle sum property and exterior angle property of a triangle.

Interior Angle Sum Property of Triangle

Theorem: The sum of interior angles of a triangle is 180° or two right angles (2x 90° )

Given: Consider a triangle ABC.

To Prove: ∠A + ∠B + ∠C = 180°

Construction: Draw a line PQ parallel to side BC of the given triangle and passing through point A.

Proof: Since PQ is a straight line, From linear pair it can be concluded that:

∠1 + ∠2+ ∠3 = 180° ………(1)

Since, PQ || BC and AB, AC are transversals

Therefore, ∠3 = ∠ACB (a pair of alternate angles)

Also, ∠1 = ∠ABC (a pair of alternate angles)

Substituting the value of ∠3 and ∠1 in equation (1),

∠ABC + ∠BAC + ∠ACB = 180°

⇒ ∠A + ∠B + ∠C = 180° = 2 x 90° = 2 right angles

Thus, the sum of the interior angles of a triangle is 180°.

 

Exterior Angle Property of Triangle

Theorem: If any one side of a triangle is produced then the exterior angle so formed is equal to the sum of two interior opposite angles.

Given: Consider a triangle ABC whose side BC is extended D, to form exterior angle ∠ACD.

To Prove: ∠ACD = ∠BAC + ∠ABC or, ∠4 = ∠1 + ∠2

Proof: ∠3 and ∠4 form a linear pair because they represent the adjacent angles on a straight line.

Thus, ∠3 + ∠4 = 180° ……….(2)

Also, from the interior angle sum property of triangle, it follows from the above triangle that:

∠1 + ∠2 + ∠3 = 180° ……….(3)

From equation (2) and (3) it follows that:

  ∠4 = ∠1 + ∠2

⇒ ∠ACD = ∠BAC + ∠ABC

Thus, the exterior angle of a triangle is equal to the sum of its opposite interior angles.

Note:

Following are some important points related to angles of a triangle:

1. Each angle of an equilateral triangle is 60°. 

2. The angles opposite to equal sides of an isosceles triangle are equal.

3. A triangle can not have more than one right angle or more than one obtuse angle.

4. In the right-angled triangle, the sum of two acute angles is 90°.

5. The angle opposite to the longer side is larger and vice-versa.

Angle Sum Property of A Triangle‌

A triangle is the smallest polygon. It has three interior angles on each of its vertices. Triangles are classified on the basis of

Interior angles as an acute-angled triangle, obtuse-angled triangle and right-angled triangle.

Length of sides as an equilateral triangle, isosceles triangle and scalene triangle.

A common property of all kinds of triangles is the angle sum property. The angle sum property of triangles is 180°. This means that the sum of all the interior angles of a triangle is equal to 180°. This property is useful in calculating the missing angle in a triangle or to verify whether the given shape is a triangle or not. It is also frequently used to calculate the exterior angles of a triangle when interior angles are given. For example,

In a given triangle ABC,

∠ABC + ∠ACB + ∠CAB = 180°

When two interior angles of a triangle are known, it is possible to determine the third angle using the Triangle Angle Sum Theorem. To find the third unknown angle of a triangle, subtract the sum of the two known angles from 180 degrees.

Let’s take a look at a few example problems:

Example 1

Triangle ABC is such that, ∠A = 38° and ∠B = 134°. Calculate ∠C.

Solution

By Triangle Angle Sum Theorem, we have;

∠A + ∠B + ∠C = 180°

⇒ 38° + 134° + ∠Z = 180°

⇒ 172° + ∠C = 180°

Subtract both sides by 172°

⇒ 172° – 172° + ∠C = 180° – 172°

Therefore, ∠C = 8°

 

Solved Examples:

1. Two angles of a triangle are of measure 600 and 450. Find the measure of the third angle.

Solution: Let the third angle be ∠A and the ∠B = 600 and ∠C = 450. Then, 

By interior angle sum property of triangles,

    ∠A + ∠B + ∠C = 1800

⇒ ∠A + 600 + 450 = 1800

⇒ ∠A + 1050 = 1800

⇒ ∠A = 180 -1050

⇒ ∠A = 750

So, the measure of the third angle of the given triangle is 750.

2. If the angles of a triangle are in the ratio 2:3:4, determine the three angles.

Solution: Let the ratio be x.

So, the angles are 2x, 3x and 4x.

By interior angle sum property of triangle,

⇒ 2x + 3x + 4x =1800

⇒ 9x = 1800

⇒ x = 1800/ 9

⇒ x = 200

The three angles are:

2x = 2(200) = 400

3x = 3(200) = 600

4x = 4(200) = 800

So, the three angles of the triangle are 400, 600 and 800 respectively.

3. Find the values of x and y in the following triangle.

Solution: Using exterior angle property of triangle,

x + 50° = 92° (sum of opposite interior angles = exterior angle)

⇒ x = 92° – 50°

⇒ x = 42°

And, 

y + 92° = 180° (interior angle + adjacent exterior angle = 180°.)

⇒ y = 180° – 92° 

⇒ y = 88°

So, the required values of x and y are 42° and 88° respectively

This chapter Application of derivatives mainly features a set of topics just like the rate of change of quantities, Increasing and decreasing functions, Tangents and normals, Approximations, Maxima and minima, and lots more. We are going to discuss the important concepts of the chapter application of derivatives.

1. A function f is said to be

1. Increasing only if on an interval of (a, b) if x₁ < x₂ 

2. Decreasing on (a,b) if x₁ < x₂

3. Let the constant (a, b), if f (x) equals c for all x ∈ (a, b), where c is the constant.

2. First Derivative Test.

3. Second Derivative Test.

The main topics which are covered for the NCERT Solutions for Application of Derivatives Class 12 notes, Chapter 6 are:

NCERT Solutions for Application of Derivatives Class 12 Notes Mathematics Chapter 6

Maximum and Minimum Value

Let f be the function which is defined on an interval I. So,

1. f is claimed to possess a maximum value in I, if there exists some extent c in I such: f(c) > f(x), ∀ x ∈ I. The number f(c) is named the utmost value of f in I and therefore the point c is named to some extent a maximum value of f in I.

2. f is said to be having a minimum value in I, and if there exists a point c in I so f(c) < f(x), ∀ x ∈ I. The number f(c) is named the minimum value of f in I and therefore the point c is named to some extent the minimum value of f in I.

3. f is claimed to possess an extreme value in I, if there exists some extent c in I such f(c) is either a maximum value or a minimum value of f in I. The number f(c) is named an extreme value off in I and therefore the point c is named an extreme.

Important Points of Applications of Derivatives

1. Through the graphs, we will even find the maximum/minimum value of a function to some extent at which it’s not even differentiable.

2. Every monotonic function makes sure that its maximum/minimum value is at the endpoints of the domain of the definition of the function.

Every continuous function on a bounded interval features a maximum and a minimum value.

Let f be a function which is defined on an unbounded interval which is I. Suppose cel is any point. If f has local maxima or local minima at x = c, then either f'(c) = 0 or f isn’t differentiable at c.

Critical Point: to some extent c within the domain of a function f at which either f'(c) = 0 or f isn’t differentiable, is named a juncture of f.

First Derivative Test: Let f be a function defined on an unbounded interval which is I and f be the continuous of a juncture c in I. So, if f'(x) changes sign from positive to negative as x increases through c, then c may be a point of local maxima.

if f'(x) changes sign from negative to positive as x increases through c, then c may be a point of local minima.

if f'(x) doesn’t change sign as x increases through c, then c is neither some extent of local maxima nor some extent of local minima. Such some extent is named some extent of inflection.

Second Derivative Test: Let f(x) be a function that is defined on an interval known as I and c ∈ I. Let f be two times differentiable at c. So,

1. x = c is a point for the local maxima, if f'(c) equals 0 and f”(c) < 0.

2. x = c is a point for the local minima, if f'(c) equals 0 and f”(c) > 0.

3. The test will fail if f'(c) equals 0 and f”(c) equals 0.

A solid surface produced by a line moving parallel to a fixed line, while its end describes a closed figure in a plane is called a cylinder. A cylinder is the limiting case of a prism. This old-fashioned view is utilized in fundamental utilizations of geometry, yet the unpredictable numerical perspective has moved to the endless curvilinear surface and this is the manner by which a cylinder is currently all around characterized in a few present-day parts of geometry and topology. The shift in the straightforward meaning has created some uncertainty with terminology.

If a line is perpendicular to the base, the cylinder is called a Right cylinder, otherwise, it is called an oblique cylinder. The line joining the centers of the bases is called the axis of the cylinder. A hollow cylinder is a cylinder which is vacant from inside and has some difference between the internal and external radius. 

Parts of a Cylinder:

Base and Side

A cylinder is a solid that is common to see in regular day to day existence, for example, a straw. In the event that you dismantle it, you discover it has two closures, called bases, that are normally roundabout. The bases are consistent and parallel to one another. If you somehow managed to ‘unroll’ the cylinder you would locate the side is really a rectangle shape when straightened out.

Height

The height h is the perpendicular distance between the 2 bases. It is important to use the perpendicular height (‘altitude’) when we calculate the volume of a slanted cylinder.

Radius

The radius r of a cylinder is the radius of the base. If you are given the diameter instead, remember to take half of it.

Axis

A line joining the center of each of the 2 bases.

Real Life Examples:

1. Tubes

2. Circular Buildings

3. Straws

Deducing the Formulae of Areas of a Hollow Cylinder:

If R is the outer radius of the cylinder and r is the inner radius of the cylinder, then

(i) Volume (the solid portion) = Volume of external cylinder-volume of internal cylinder

= [pi R^{2} h – pi r^{2} h]

= [pi (R^{2} – r^{2}) h]

(ii) Lateral surface area = External surface area of a cylinder + Internal surface area of a cylinder

= [2 pi Rh + 2 pi rh]

= [2 pi h(R + r)]

(iii) Total surface area = Lateral surface area + Areas of solid bases

= [2 pi h(R + r) + 2 pi (R^{2} – r^{2})]

Example 1:

Find the weight, lateral surface area and total surface area of a steel pipe whose interior and exterior diameters measure 15cm and 17cm respectively, and length 10m; one cubic cm of iron weighing 0.8gm.

Solution:

Here d = 15cm r = 7.5cm

D = 17 cm R = 8.5 cm

h = 10 m = 1000 cm

[text{Volume} = pi (R^{2} – r^{2})h]

= [pi (72.25 – 56.25)1000]

= [50265.48 cm^{3}].

Weight = Volume x density = 50265.48 x 0.8 = 40212.39 gms

[text{Lateral surface area} = 2 pi (R + r)h]

= [2 pi (8.5 + 7.5)1000]

= [2 pi times 16 times 1000]

= [100530.96 cm^{2}].

[text{Total surface area of the pipe} = text{Lateral surface area of pipe} + text{Area of bases}]

= [100530.96 + 100.53]

= [100631.49 cm^{2}].

Example 2:

A hollow cylinder copper pipe is 21dm long. Its outer diameter and inner diameter are 10cm and 6cm respectively. Find the volume of copper used in manufacturing the pipe.

Solution:

Given that:

[text{The height of the cylindrical pipe is h} = 21 dm = 210 cm]

[text{Thus, External radius}, R = frac{10}{2} = 5cm]

[text{Internal radius}, r = frac{6}{2} = 3 cm]

The volume of the copper used in manufacturing the pipe 

= [text{Volume of external cylinder} – text{volume of an internal cylinder}]

= [pi R^{2} h – pi r^{2} h]

= [pi (R^{2} – r^{2})h]

= [frac{22}{7} [5^{2} – 3^{2}] times 210 = frac{22}{7} times 16 times 210]

= [22 times 16 times 30]

= [10560 cu.cm]

Example 3:

The inner radius of a circular well is 2.1 m and its depth is 21 m. Find the cost of plastering the inner surface of a circular well at the rate of Rs. 40 per m².

Solution:

Given: Radius of the circular well (r) = 2.1 m, depth (h) = 21 m

Here we need to plaster the inner surface of the well which is the sum of Curved surface area and Area of the base.

[text{Area to be plastered} = text{Curved surface area} + text{Area of base}]

= [2 pi rh + pi r^{2}]

= [2 times frac{22}{7} times 2.1 times 21 + frac{22}{7} times (2.1)^{2}]

= [277.2 + 13.86]

= [291.06 m^{2}]

[text{Cost of} 1 m^{2} text{of well} = Rs. 40]

Cost of 291.06 m2 of well = Rs. (40 × 291.06) = Rs. 11642.40

Test Sample:

Example:

The lateral surface area of a hollow tube is 4224 cm². Later on it has been cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of a rectangular sheet?

(Answer=322cm.)

Example:

The total surface area of a hollow metal tube, open at both ends of external radius 8 cm and height 10 cm is 338π cm2. Taking r to be the inner radius, provide an equation in r and use it to state the thickness of the metal in the cylinder.

(Answer=3cm)

Example:

The total surface area of a hollow ceramic cylinder which is open from both the sides is 4620 sq. cm, the base ring area is 115.5 sq. cm and height 7 cm. Find the thickness of the ceramic cylinder.

(Answer=7/19cm)

Example:

Find the cost of plastering the inner surface of a well at the rate of Rs 30 per m2, if the inner diameter of the well is 2.8 m and its depth is 14 m.

(Answer= Rs 3880.80)

Example:

A hollow cylinder copper pipe is 21cm long. It’s the outer diameter and the inner diameter is 10cm and 6cm respectively. Find the volume of copper used in manufacturing the pipe.

(Answer=340 cubic. cm)

A sector is much the same as a “pie” or a “pizza slice”. It typically involves a province (sector) encompassed by 2 Radii and 1 Arc lying between the radii. To better understand what a sector of a circle is, here is a specimen from the real-life examples of food. A variety of foods that we eat resembles the sector of a circle, typically because they replicate circles. 

 

Find the Sector in the Circle

A circle consists of an interior, its area. Divide the circle into pieces and you will get a sector. Similarly, cut the cake into Slices and you will get the sector of a circle.

To ace over the calculation of area of sector, understanding the anatomy of the circle is the key. 

A Midpoint – It is the most crucial one as it is the connecting dot positioned right in the circle’s centre.

Radius – A line that runs through the midpoint to the exterior of the circle

Diameter – A line extending across the circle and passes through its midpoint. 

Take Notice of Diameter – it is actually two radii combined. Therefore, a circle’s radius is invariably half the length of its diameter and its diameter is always double the length of its radius.

A semicircle (half circle) and a Quadrant (Quarter of a circle) are two main types of Sector:

Besides, There are primarily two “portions” of a circle: Of which, the “pizza” slice is known as the Sector. Whereas, a cut carried off from between two points on the circle by a “chord” is called a Segment. This means Chunks cut off by joining any two points on the circle are segments. 

Seeing that, both sectors and segments are part of a circle’s interior, both have areas. We can measure their area using formulas.

Easy Formulas to Calculate Area of a Sector of a Circle

1. Calculating the Area of Sector of a Circle Using Degrees

A whole of a circle surrounds 360°, thus the ratio of the sector’s angle calculation to 360° is directly proportional to the fraction of the circle’s area is computed. Hence, for the area of the sector of a circle, it’s the median angle of 360°, times the area of the circle. The sector of a circle that subtends an angle of 180 degrees at the centre is called a semicircle. For instance, if the sector’s angle happens to measure 180°, and the two radiuses creating it are 10 inches, you would divide 180 by 360 (180/360) or 1/2 to obtain the portion of the circle.

In numeric terms, the area of a sector of a circle formula will be = [frac {theta}{360} times pi r^{2}]

Hence, the area of the semicircle will be = [frac {1}{2} times pi r^{2}]

This is because the angle is 180 degrees.

Solved Example

Question: Given is a circle with a radius of 2 cm and ∠ A = 30°. Find the area of the sector of the circle below? 

Solution: Area of circle = πr2 =  π 22 = 4π

 

Total degrees in a circle = 360°

 

Given that the central angle is 30 degrees and the radius is 2cm,  

 

Therefore, 30° slice = [frac {30}{360}] fraction of circle. 

 

= [frac {30}{360} times pi r^{2}]. 

 

 Area of sector = [frac {theta}{360}]* Total Area 

 

                           = [frac {theta}{360} times pi r^{2}]

 

                           =  [frac {1}{12} times frac {22}{7} times 4]

 

                          = 1.047 square cm                                      

2. Calculating the Area of Sector of a Circle Using Radians

If the question is available with radians as a replacement for degrees to calculate the area of sector angle, the usual method of finding the sector’s area remains the same. Radians are units that are utilized for computing angles. A circle on a whole circumscribes 2 π radians, so the portion of 2 π is the fraction of the circle’s area that is being measured.

To put it simply, if the sector’s angle is [frac {pi}{2}] radians, the portion of the circle to be measured will be [frac {pi}{2}] divided by 2π, or [frac {1}{4}] of the circle. 

So, to find the area, multiply the circle’s area by the fraction of the circle that is being dealt with. Just make use of radians instead of degrees. The sector of a circle formula in radians is:  

A =  [frac{text{angle in radian}}{pi} times pi r^{2}]  

For example, if a sector contains an angle of [frac {pi}{3}] at the centre. Then the area of the sector would be  [frac {pi}{3}]

= [frac{frac{pi}{3}}{pi} times pi r^{2}] 

 = [frac {1}{3} times pi r^{2}]

3. Calculating the Area of Sector Using the Known Portions of a Circle

In cases where the portion of a circle is known, don’t divide degrees or radians by any value.

 

For example, if the known sector is 1/4 of a circle, then just multiply the formula for the area of a circle by ¼, and you are good to go to find the area of the sector. 

 

Thankfully, this sector of a circle formula would work with any stated fraction of a circle. Even if the known portion is 1/120 of a circle, simply multiply the formula for the area of a circle by 1/120.

 

So, when given the portion of the circle, the sector of a circle formula is:

 

A = (portion of the circle) (π x r2)

 

For example, if we have to find the area of the quadrant of a circle, then it is already known to us that the angle subtended by the quadrant at the centre of the circle is always 90 degrees and it divides the circle into four parts of equal area.

 

Hence, we can also say that the area of a single quadrant of a circle is ¼ of the area of the circle. 

 

Area of Quadrant =[frac {90}{360} times pi r^{2}] 

 

= [frac {1}{4} times pi r^{2}] 

 

So if the radius of a circle is 10 cm then the area of its one of the quadrant would be :

 

= [frac {1}{4} times pi times 100]

 

= 78.57 square cm.

 

If the angle at the centre that is θ wou
ld be in radians, area of the sector of a circle = [frac {1}{2} times r^{2} theta], where,

Here, θ will be the angle subtended at the centre, given in radians and r is the radius of the circle.

It should be noted that semicircles and quadrants are special types of sectors of a circle. They made an angle of 180° and 90° respectively with the centre.

Area of Sector Using Degrees

Now let us discuss the area of the sector of a circle whose angle is given to you in degrees with the help of the example given here.

Example: Determine the area of a sector if the radius of the circle is 8 units, and the angle subtended at the centre = [frac {pi}{3}]. 

Solution: Given, radius = 8 units; Angle measure (θ)= [frac {pi}{3}]

The area of the given sector can be calculated with the given formula, 

[A = (frac {text{sector angle}}{360}) times (pi times r^{2}) ]

Hence, Area of sector would be  = [(frac {theta}{360}) times pi r^{2}]

When we substitute the given values  we get the Area of the sector  

= [frac{frac{pi}{3}}{360} times frac {22}{7} times 64]

=12π

Hence, the area of the given sector in radians is expressed as 12π square units

Sequence means an order.  Now, the point is what is a sequence and how is it related to the subject of Mathematics. An arithmetic sequence is about numbers that add or subtract.  The geometric progression goes from one term to another and multiplies or divides. The result is a shared value. Ref the figure below, in the arithmetic sequence the difference (d) is always a standard value 7. In the geometric series, the common value is always 2. An arithmetic sequence is about addition (or subtraction) in a set order. A geometric sequence is about multiplying (or division) in a set order.

Arithmetic Sequence Formula

Let us understand the Arithmetic sequence formula. In the arithmetic sequence, one term goes to the next term by always adding—for example, 1, 2, 3, 4, 5 ….10, and so on. Here, in this sequence, each number moves to the second number by adding (or subtracting 1). Let us take some examples to understand better. 2, 5, 8, 11, 14….is arithmetic sequence as each step adds 3. The same holds for a reverse order (in subtraction). E.g. 7, 3, -1, -5 …is an arithmetic sequence as each step subtracts 4.

It is essential to note that the number that is added or subtracted at each level of an arithmetic sequence is called as the difference (d). The reason is that if you add or subtract (also known as finding the difference), you always get the same common value. Ref fig 2 below

Geometric Sequence Formula

A geometric sequence or geometric series is a geometric order. We obtain results by multiplying the terms of a geometric sequence. In simple words, a geometric sequence moves from one term to the next by always multiplying (or by division) by the same common value or number. For example, 2, 4, 8, 16, 32 … is a geometric series. The reason being that each step multiplies by two. Let us take another example, 81, 27, 9, 3, 1 …it is a geometric sequence as each step divides (or multiplies) by the number 3.

It is essential to note that the common number that multiplies or divides at each step of a geometric sequence is called the ratio r. It is due to the fact that if you divide or find the ratio of the successive terms, you get a common or standard ratio. Ref Fig.3 below.

be added soon

Arithmetic Geometric Progression 

An arithmetic progression is a series or sequence of numbers in which each term is derived from the next term, by adding or subtracting a fixed or common number called the common difference d. For instance, the series, 9, 6, 3, 0,-3 and so on, is an arithmetic progression with -3 as the standard difference. The progression -3, 0, 3, 6, 9 is an example of Arithmetic Progression (AP) that has three as the common difference d.

The established form of AP is a, a + d, a + 2d, a + 3d and ….so on. The nth term will be

an = a + (n -1)d  

A geometric progression is a series in which each term is obtained by a multiplication or division of the next term, by a fixed or common number. For instance, the series 8, 4, -2, 1, -1/2 … is a Geometric Progression (GP) for which -1/2 is the common ratio.  

The established form of GP is a, ar, ar2, ar3, ar4….and so on. The nth term will be   an = ar (n-1)  

Example -1

Work out the common difference and the next term of the following series: 

3, 11, 19, 27, 35 … 

We have to find the common difference d. you can pick up any pair. Let us start with subtractions. 

11 – 3 = 8 

19 -11 = 8 

27 -19 = 8 

35 – 27 = 8 

Throughout, the difference is 8, so the common difference is 8.

We have 5 terms. We have to find the next or the 6th term. We can find out by adding the common difference to the fifth term, 35 + 8 = 43. 

sixth term = 43 and common difference =8

Example -2 

Find the common ratio and the 7th term of the following sequence 

2/9, 2/3, 2, 6, 18. 

We will take 6/2 = 3 and 18/6 = 3, the common ratio is 3, so r = 3  

We have 5 terms, we have to find the 6th, then the 7th term. 

So, a6 = 18 x 3 = 54 

 a7 = 54 x 3 = 162, the answer is common ratio r = 3 and seventh term = 162.