{"id":5193,"date":"2023-05-19T11:48:20","date_gmt":"2023-05-19T06:18:20","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=5193"},"modified":"2023-05-20T17:49:08","modified_gmt":"2023-05-20T12:19:08","slug":"ap-board-9th-class-maths-solutions-chapter-15-ex-15-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/ap-board-9th-class-maths-solutions-chapter-15-ex-15-3\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 15 Proofs in Mathematics Ex 15.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 15th Lesson Proofs in Mathematics Exercise 15.3<\/h2>\n

\"AP<\/p>\n

Question 1.
\nTake any three consecutive odd numbers and find their product, for example 1 \u00d7 3 \u00d7 5 = 15;
\n3 \u00d7 5 \u00d7 7 = 105: 5 \u00d7 7 \u00d7 9 = ……………
\nii) Take any three consecutive even numbers and add them, say,
\n2 + 4 + 6 = 12; 4 + 6 + 8 = 18:
\n6 + 8 + 10 = 24; 8 + 10 + 12 = 30 ….
\nso on. Is there any pattern you can guess in these sums ? What can von conjecture about them ?
\nSolution:
\ni) 1 \u00d7 3 \u00d7 5 = 15
\n3 \u00d7 5\u00d7 7 = 105
\n5 \u00d7 7 \u00d7 9 = 315
\n7 \u00d7 9 \u00d7 11 = 693<\/p>\n

    \n
  1. The product of any three consecutive odd numbers is odd.<\/li>\n
  2. The product of any three consecutive odd numbers is divisible by \u20193′.<\/li>\n
  3. 2 + 4 + 6 = 12; 4 + 6 + 8= 18;
    \n6 + 8 + 10 = 24; 8 + 10 + 12 = 30<\/li>\n
  4. The sum of any three consecutive even numbers is even.<\/li>\n
  5. The sum of any three consecutive even numbers is divisible by 6<\/li>\n
  6. The sum of any three consecutive even numbers is a multiple of 6.<\/li>\n<\/ol>\n

    \"AP<\/p>\n

    Question 2.
    \nGo back to Pascal’s triangle.
    \nLine-1: 1=11\u00b0
    \nLine-2: 11 = 111<\/sup>
    \nLine-3 : 121 = 112<\/sup>
    \nMake a conjecture about line – 4 and line – 5.
    \n\"AP
    \nDoes your conjecture hold ? Does your conjecture hold for line – 6 too ?
    \nSolution:
    \nLine-4 : 1331 = 113<\/sup>
    \nLine-5 : 14641 = 114<\/sup>
    \nLine – 6 : 115<\/sup>
    \n\u2234 Line – n = 11n-1<\/sup>
    \nYes the conjecture holds good for line – 6 too.<\/p>\n

    \"AP<\/p>\n

    Question 3.
    \nLook at the following pattern.
    \ni) 28 = 22<\/sup> \u00d7 71<\/sup>;
    \nTotal number of factors
    \n(2+ 1)(1 + 1) = 3 \u00d7 2 = 6
    \n28 is divisible by 6 factors i.e.,
    \n1, 2, 4, 7, 14, 28.
    \nii) 30 = 21<\/sup> \u00d7 31<\/sup> \u00d7 51<\/sup>, Total number of .
    \nfactors (1 + 1) (1 + 1) (1 + 1) = 2 \u00d7 2 \u00d7 2 = 8
    \n30 Is divisible by 8 factors i.e., 1, 2, 3, 5, 6, 10, 15 and 30
    \nFind the pattern.
    \n[Hint : Product of every prime base exponent +1)
    \nSolution:
    \n24 = 23<\/sup> \u00d7 31<\/sup>
    \n24 has (3+1) (1 + 1) = 4 \u00d7 2 = 8 factors
    \n[1, 2, 3, 4, 6, 8, 12 and 24]
    \n36 = 22<\/sup> \u00d7 32<\/sup>
    \nNumber of factors = (2 + 1) (2 + 1)
    \n3 \u00d7 3 = 9 [ 1, 2, 3. 4, 6, 9, 12, 18 and 36]
    \nIf N = ap<\/sup>. bq<\/sup> . cr<\/sup>…….. where N is a natural number.
    \na. b, c … are primes and p, q, r ……. are positive integers then, the number of factors of N =(p- 1)(q+ 1)(r + 1)<\/p>\n

    \"AP<\/p>\n

    Question 4.
    \nLook at the following pattern :
    \n12<\/sup> = 1
    \n112<\/sup> = 121
    \n1112<\/sup> = 12321;
    \n11112<\/sup> = 1234321
    \n111112<\/sup> = 123454321
    \nMake a conjecture about each of the following
    \n1111112<\/sup> =
    \n11111112<\/sup> =
    \nCheck if your conjecture is true.
    \nSolution:
    \n1111112<\/sup> = 12345654321
    \n11111112<\/sup> = 1234567654321
    \n(111………. n-times)2<\/sup> = (123 … (n- 1) n (n – 1) (n – 2) 1)
    \nThe conjecture is true.<\/p>\n

    \"AP<\/p>\n

    Question 5.
    \nList five axioms (postulates) used in text book.
    \nSolution:<\/p>\n

      \n
    1. Things which are equal to the same things are equal to one another.<\/li>\n
    2. If equals are added to equals, the sums are equal.<\/li>\n
    3. If equals are subtracted from equals, the differences are equal.<\/li>\n
    4. When a pair of parallel lines are in-tersected by a transversal, the pairs of corresponding angles are equal.<\/li>\n
    5. There passes infinitely many lines through a given point.<\/li>\n<\/ol>\n

      \"AP<\/p>\n

      Question 6.
      \nIn a polynomial p(x) = x2<\/sup> + 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 ?
      \nSolution:
      \np(x) = x2<\/sup> + x + 41
      \np(0) = 02<\/sup> + 0 + 41 = 41 – is a prime
      \np(1) = 12<\/sup> + 1 + 41 = 43 – is a prime
      \np(2) = 22<\/sup> + 2 + 41 = 47 – is a prime
      \np(3) = 32<\/sup> + 3 + 41 = 53 – is a prime
      \np(41) = 412<\/sup> + 41 + 41
      \n= 41(41 + 1 + 1)
      \n= 41 x 43 is not a prime.
      \n\u2234 p(x) = x2<\/sup>+ x + 41 is not a prime for all x.
      \n\u2234 The conjecture “p(x) = x2<\/sup> + x + 41 is a prime” is false.<\/p>\n","protected":false},"excerpt":{"rendered":"

      AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 15 Proofs in Mathematics Ex 15.3 Textbook Questions and Answers. 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 \u00d7 3 \u00d7 5 = 15; 3 … Read more<\/a><\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/5193"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=5193"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/5193\/revisions"}],"predecessor-version":[{"id":40600,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/5193\/revisions\/40600"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=5193"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=5193"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=5193"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}