AP Board Class 9 Maths Chapter 15 Proofs in Mathematics Ex 15.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 15 Proofs in Mathematics Ex 15.3 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

AP Board Class 9 Maths Chapter 15 Proofs in Mathematics Ex 15.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 9th Maths Chapter 15 Proofs in Mathematics Ex 15.3 Book Answers

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Andhra Pradesh State Board Class 9th Maths Chapter 15 Proofs in Mathematics Ex 15.3 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths
Chapters	Maths Chapter 15 Proofs in Mathematics Ex 15.3
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AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.3

Question 1.
Take any three consecutive odd numbers and find their product, for example 1 × 3 × 5 = 15;
3 × 5 × 7 = 105: 5 × 7 × 9 = ……………
ii) Take any three consecutive even numbers and add them, say,
2 + 4 + 6 = 12; 4 + 6 + 8 = 18:
6 + 8 + 10 = 24; 8 + 10 + 12 = 30 ….
so on. Is there any pattern you can guess in these sums ? What can von conjecture about them ?
Solution:
i) 1 × 3 × 5 = 15
3 × 5× 7 = 105
5 × 7 × 9 = 315
7 × 9 × 11 = 693

The product of any three consecutive odd numbers is odd.
The product of any three consecutive odd numbers is divisible by ’3′.
2 + 4 + 6 = 12; 4 + 6 + 8= 18;
6 + 8 + 10 = 24; 8 + 10 + 12 = 30
The sum of any three consecutive even numbers is even.
The sum of any three consecutive even numbers is divisible by 6
The sum of any three consecutive even numbers is a multiple of 6.

Question 2.
Go back to Pascal’s triangle.
Line-1: 1=11°
Line-2: 11 = 11¹
Line-3 : 121 = 11²
Make a conjecture about line – 4 and line – 5.

Does your conjecture hold ? Does your conjecture hold for line – 6 too ?
Solution:
Line-4 : 1331 = 11³
Line-5 : 14641 = 11⁴
Line – 6 : 11⁵
∴ Line – n = 11^n-1
Yes the conjecture holds good for line – 6 too.

Question 3.
Look at the following pattern.
i) 28 = 2² × 7¹;
Total number of factors
(2+ 1)(1 + 1) = 3 × 2 = 6
28 is divisible by 6 factors i.e.,
1, 2, 4, 7, 14, 28.
ii) 30 = 2¹ × 3¹ × 5¹, Total number of .
factors (1 + 1) (1 + 1) (1 + 1) = 2 × 2 × 2 = 8
30 Is divisible by 8 factors i.e., 1, 2, 3, 5, 6, 10, 15 and 30
Find the pattern.
[Hint : Product of every prime base exponent +1)
Solution:
24 = 2³ × 3¹
24 has (3+1) (1 + 1) = 4 × 2 = 8 factors
[1, 2, 3, 4, 6, 8, 12 and 24]
36 = 2² × 3²
Number of factors = (2 + 1) (2 + 1)
3 × 3 = 9 [ 1, 2, 3. 4, 6, 9, 12, 18 and 36]
If N = a^p. b^q . c^r…….. where N is a natural number.
a. b, c … are primes and p, q, r ……. are positive integers then, the number of factors of N =(p- 1)(q+ 1)(r + 1)

Question 4.
Look at the following pattern :
1² = 1
11² = 121
111² = 12321;
1111² = 1234321
11111² = 123454321
Make a conjecture about each of the following
111111² =
1111111² =
Check if your conjecture is true.
Solution:
111111² = 12345654321
1111111² = 1234567654321
(111………. n-times)² = (123 … (n- 1) n (n – 1) (n – 2) 1)
The conjecture is true.

Question 5.
List five axioms (postulates) used in text book.
Solution:

Things which are equal to the same things are equal to one another.
If equals are added to equals, the sums are equal.
If equals are subtracted from equals, the differences are equal.
When a pair of parallel lines are in-tersected by a transversal, the pairs of corresponding angles are equal.
There passes infinitely many lines through a given point.

Question 6.
In a polynomial p(x) = x² + x + 41 put different values of x and find p(x). Can you conclude after putting different values of x that p(x) is prime for all. Is ‘x’ an element of N ? Put x = 41 in p(x). Now what do you find ?
Solution:
p(x) = x² + x + 41
p(0) = 0² + 0 + 41 = 41 – is a prime
p(1) = 1² + 1 + 41 = 43 – is a prime
p(2) = 2² + 2 + 41 = 47 – is a prime
p(3) = 3² + 3 + 41 = 53 – is a prime
p(41) = 41² + 41 + 41
= 41(41 + 1 + 1)
= 41 x 43 is not a prime.
∴ p(x) = x²+ x + 41 is not a prime for all x.
∴ The conjecture “p(x) = x² + x + 41 is a prime” is false.

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