Learn Unit-V: Mathematical Reasoning with Free Lessons & Tips

                                            Geometric Progression Example.

Q: In a GP,the third term is 24 and the 6^th Term is 192. Find the 10^th  term?

                                                                  Solution:

We know that the nth  term of an GP is given by:

                                   (a*r^(n-1)).      ….. (1)   

Now, as per the question,the third term is 24,therefore,

                                          24 = a*r^(3-1)      …(2)

And the 6^th term is 192,therefore,

• 192 = a*r^(5-1) …(3)

Therefore,

                             a* r^(5-1) / a*r^(3-1) = 192/24

• r^3 = 8 ==>  r = 2

Thus,we have the common ratio as 2. Now, we will find the first term i.e. a.

Substituting r = 2 in …(2), we have, 4a = 24 (because r^2 = 4) from which a = 6.

Now, we have both the first term as well as the common ratio. Therefore, now to get the 10^th term we simply substitute the values of a and r in a*r^(10-1)and obtain

                                            6*(2^9) which comes out to be 3072.

Thus, we have, as the 10^th term,3072.

Math Phobia or math anxiety is a very real problem that many young students and even adult professionals face.

What actually is Math Phobia? It is the doubt on one's mathematical abilities.

How does it begin? Many underlying reasons are there to fear and hate maths. It could be because of a substandard teacher at a young age. It could be because of parents' math anxiety which they subconsciously transferred in their child. The pressure of exams and tests, maths being taught as a right or wrong subject instead of a creative field of study are more reasons why Maths is the most feared subject in the world.

How to get better and love maths? Teachers and parents need to motivate and help students stay positive and enthusiastic about learning Math. One way is to give them open-ended problems to solve and discuss. If you realise the uses and applications of different fields of Maths in everyday lifestyle like percentages, estimations and fractions when shopping, measuring and fractions in cooking, decimals in financial transactions, trigonometry in flying plane and construction and so much more.

DO maths to LEARN maths. Solve as many mathematical problems, experiment on understanding, think before asking.

Keep going, keep learning.

Natural Numbers: The set of counting numbers is called natural numbers. It is denoted by N, where N = {1,2,3, …..}

Whole Numbers: when zero is included in the set of natural numbers, then it is a set of whole numbers. It is denoted by W, where W = {0,1,2, …..}

Integers: When in the set of whole numbers, natural numbers with the negative sign are included, then it becomes a set of integers. It is denoted by Z, where Z = {-…. , -2, -1, 0, 1, 2, …..}

Integers can be divided into negative and positive integers.

Prime Numbers: The natural numbers which have no factors other than 1 and itself are called prime numbers.

Co-prime Numbers: Two numbers that have no common factor except one are called co-prime numbers. Example: 11 and 18 etc.

Rational Numbers: The numbers that can be expressed in the form of p/q where p and q are integers and co-prime are called rational numbers. It is denoted by Q. Rationalnumbers can be positive or negative.

Irrational Numbers : The numbers which are not rational numbers, are called irrational numbers. or π = 3.141592 is an irrational number.

Real Numbers : Set of all rational numbers as well as irrational numbers are called real numbers. The square of all real numbers is positive.

Some important points on numbers :

• 2 is the only even prime number.
• Number 1 is neither divisible nor prime.
• Two consecutive odd prime numbers are called prime pair.
• All natural numbers are whole, rational, integer and real.
• Fractions are rational.
• 0 is neither negative(-) nor positive(+) number.
• Dividing any number by 0 gives infinity () whereas dividing 0 by any number gives 0.
• The sum and product of two rational numbers is always a rational number.
• The product and sum of a rational number and irrational number is always an irrational number.
• There can be infinite numer of rational and irrational numbers between two rational numbers and two irrational numbers.

In number theory, we deal only with integers and their properties. Many of you may have heard about this branch of mathematics but may not know much about it. Here I am going to introduce you to some of the basic concepts of this branch of mathematics. So, let's begin :)

One of the primary concepts essential for learning number theory is the divisibility of integers. The first of this list is Euclid's division algorithm which states that:

Given integers a and b with b>0, there exist unique integers q and r such that a=qb+r.

Well, this is what we have learnt while learning division—considering an integer a and an integer b if we divide a by b. we get a quotient and a remainder. In the above theorem, q is the quotient and r is the remainder. But the exciting thing is that such a representation of an integer is always unique, which is quite apparent. 

Let's see how can we use this property to solve some interesting problems. But before that, you all must be knowing that an even integer N can be written as N=2k where k is also an integer. If N is odd, it can be represented as N=2k+1.

Problem 1: For n≥1 establish that the integer M=n(7n² +5) is divisible by6, where n is an integer.

Solution: What do you think? Isn't it interesting that a number of this form will always be divisible by 6 irrespective of what value we put in place of n, if we place 1, it's true, same for 2, same for any n? But seeing the form, you cannot tell that the number will be divisible by 6.

So, let's see how we can approach the problem.

M=n(7n² +5) =n(6n² + n² +6-1)=6n³ + n³ +6n -n = (6n³⁺ 6n) + (n³ -n). Let me tell you how this helps. We needed to prove that that M is divisible by 6. So, I decided to take out those terms which have 6 as a factor. Quite clearly  (6n³⁺ 6n)  will be divisible by 6. We just need to prove that (n³ -n)  will be divisible by 6. 

Let's factorise (n³ -n). (n³ -n)=n(n+1)(n-1) . Now suppose we have a number and we divide it by 3, the possible remainders are 0,1 and 2. So we can write any integer in the following forms: 3k,3k+1,3k+2  where k is an integer. Let's see what happens if we replace n with these forms.

If n=3k. So, (n³ -n)=3k(3k+1)(3k-1). So, (n³ -n) is divisible by 3.

If n=3k+1, (n³ -n)=(3k+1)(3k+2)3k. So, (n³ -n) is divisible by 3.

If n=3k+2, (n³ -n)=(3k+2)(3k+3)(3k+1)=3(3k+2)(k+1)(3k+3) NOTE: HERE WE TOOK 3 COOMON FROM 3K+3. WE WROTE 3K+3=3*(k+1).  So, (n³ -n) is divisible by 3.

So, for all integers,  (n³ -n) is divisible by 3.

We also know that an integer is either odd or even. So, n-2k if n is even or n=2k+1 if n is odd.

If n is even, (n³ -n)=2k(2k+1)(2k-1) .So, (n³ -n) is divisible by 2.

If n is odd, (n³ -n)=2k+1(2k+2)2k. So, (n³ -n) is divisible by 2.

So, (n³ -n)  is divisible by both 2 and 3. So, (n³ -n)  is divisible by 2*3=6.

Hence, we have proved that M is divisible by 6.

  Maths is not a subject , it is a habit like our daily activities eating , sleeping and many more. As per my experience which students deal it as a game or fun as their habit , after aomestime they become champion for it . I have seen there are two type of students 1. Who take it as their habit , after some they become champion 2. Who take it as a subject , they always try to deal it as other subjects and they finds several difficulties to raise their logical power by this subhects.   So it is my best suggestion for students to death this subject as a habit , for that they should solve problems at least one hour every  day which can make them as a champion in maths....