{"id":121,"date":"2021-07-22T14:30:08","date_gmt":"2021-07-22T09:00:08","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=121"},"modified":"2021-07-22T14:42:48","modified_gmt":"2021-07-22T09:12:48","slug":"kseeb-sslc-class-10-maths-solutions-chapter-1-ex-1-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-1-ex-1-1\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1<\/strong>\u00a0are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1<\/h2>\n

KSEEB Solutions For Class 10 Maths Question 1.<\/strong>
\nIn which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
\ni) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
\nii) The amount of air present in a cylinder when a vacuum pump removes of the air remaining in the cylinder at a time.
\niii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.
\niv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.
\nSolution:
\ni) Rent for\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Rent for\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Rent for
\n1st km\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 2nd km\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a03rd km
\nRs. 15\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Rs. 8\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0Rs. 8
\nd = a2<\/sub> – a1<\/sub> = 8 – 15 = -7
\nd = a3 <\/sub>– a2<\/sub> = 8 – 8 = 0
\nHere \u2019d\u2019 is not constant.
\nHence it does not form an AP.<\/p>\n

ii) Let the amount of air present in a cylinder be ‘x’.<\/p>\n

    \n
  1. \\(\\frac{1}{4} \\text { of } 1=\\frac{1}{4}\\) Amount of air goes out
    \n\\(1-\\frac{1}{4}=\\frac{3}{4}\\) Air remaining<\/li>\n
  2. \\(\\frac{1}{4} \\text { of } \\frac{3}{4}=\\frac{3}{16}\\) Amount of air goes out
    \n\\(\\frac{3}{4}-\\frac{3}{16}=\\frac{9}{16}\\) Air remaining<\/li>\n
  3. \\(\\frac{1}{4} \\text { of } \\frac{9}{16}=\\frac{9}{64}\\) Amount of air goes out
    \n\\(\\frac{9}{16}-\\frac{9}{64}=\\frac{27}{64}\\) Air remaining<\/li>\n<\/ol>\n

    Set \\(\\frac{3}{4}, \\frac{9}{16}, \\frac{27}{64}, \\ldots .\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{2}-\\mathrm{a}_{1}=\\frac{9}{16}-\\frac{3}{4} \\quad \\mathrm{d}=\\frac{-3}{16}\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{3}-\\mathrm{a}_{2}=\\frac{27}{64}-\\frac{9}{16} \\quad \\mathrm{d}=\\frac{-9}{64}\\)
    \nHere the value of ‘d’ is not constant.
    \n\u2234 The given list of numbers does not form an A.P.<\/p>\n

    (iii) Cost of Digging for the
    \n\"KSEEB
    \nd = a2<\/sub> – a1<\/sub> = 200 – 150 = 50
    \nd = a3<\/sub> – a2<\/sub> = 250 – 200 = 50
    \nHere value of ‘d’ is constant.
    \nThis forms an AP.<\/p>\n

    iv) Amount of money deposited in the
    \nd = a2<\/sub> – a1<\/sub> = 11,664 – 10.800 = Rs. 864
    \nd = a3<\/sub> – a2<\/sub> = 12,597,12 – 11.664 = Rs. 933.12
    \nHere value of ‘d’ is not constant.
    \n\u2234 The given list of numbers does not form an AP.<\/p>\n

    KSEEB Solutions For Class 10 Maths Arithmetic Progression Question 2.<\/strong>
    \nWrite first four terms of the AP, when the first term ‘a’ and the common difference’d’ are given as follows:
    \ni) a = 10, d = 10
    \nii) a = -2, d = 0
    \niii) a = 4, d = -3
    \niv) a = -1, d = \\(\\frac{1}{2}\\)
    \nv) a = -1.25, d = 0.25
    \nSolution:
    \ni) a = 10, d = 10
    \nFirst four terms are,
    \na, a + d, a + 2d, a + 3d
    \n10, 10 + 10, 10 + 20, 10 + 30
    \n10. 20, 30, 40.<\/p>\n

    ii) a = -2, d = 0
    \nFirst four terms are,
    \na, a + d, a + 2d, a + 3d
    \n-2. -2 + 0, -2 + 0, -2 + 0
    \n-2. -2. -2, -2<\/p>\n

    iii) a = 4, d = -3
    \nFirst four terms are,
    \na, a + d, a + 2d, a + 3d
    \n4, 4 – 3, 4 – 2 \u00d7 3, 4 – 3 \u00d73
    \n4, 1, -2, -5<\/p>\n

    iv) a = -1, d = \\(\\frac{1}{2}\\)
    \nFirst four terms are,
    \na, a + d, a + 2d, a + 3d
    \n\\(-1,-1+\\frac{1}{2},-1+2 \\times \\frac{1}{2},-1+3 \\times \\frac{1}{2}\\)
    \n\\(-1,-\\frac{1}{2}, 0, \\frac{1}{2}\\)<\/p>\n

    v) a = -1.25, d = 0.25
    \nFirst four terms are,
    \na, a + d, a + 2d, a + 3d
    \n-1.25, -1.25 + 0.25, -1.25 + 2 \u00d7 0.25, -1.25 + 3 \u00d70.25
    \n-1.25, -1, -0.75, -0.50<\/p>\n

    Arithmetic Progression Exercise 1.1 KSEEB Solutions Question 3.<\/strong>
    \nFor the following APs, write the first term and the common difference :
    \n(i) 3, 1,-1,-3, ………..
    \n(ii) -5, -1, , 7, …………
    \n(iii) \\(\\frac{1}{2}, \\frac{1}{2}, \\frac{1}{2}, \\frac{1}{2}, \\ldots \\ldots\\)
    \n(iv) 0.6, 1.7, 2.8, 3.9, ………..
    \nSolution:
    \n(i) 3, 1,-1,-3
    \nFirst term, a = 3,
    \nCommon difference, d = a2<\/sub> – a1<\/sub> = 1 – 3
    \nd = -2.<\/p>\n

    (ii) -5, -1, 3, 7, ……….
    \nFirst term, a = -5,
    \nCommon difference, d = a2<\/sub> – a1<\/sub> = -1 – (-5)
    \nd = -1 + 5
    \nd = 4.<\/p>\n

    (iii) \\(\\frac{1}{2}, \\frac{1}{2}, \\frac{1}{2}, \\frac{1}{2}, \\ldots \\ldots\\)
    \nFirst term, \\(\\mathrm{a}=\\frac{1}{2}\\)
    \nCommon difference, d = a2<\/sub> – a1
    \n<\/sub>\\(=\\frac{1}{2}-\\frac{1}{2}\\)
    \nd= 0<\/p>\n

    iv) 0.6, 1.7, 2.8, 3.9
    \nFirst term, a = 0.6,
    \nCommon difference, d = a2<\/sub> – a1<\/sub> = 1.7 – 0.6
    \nd= 1.1<\/p>\n

    Arithmetic Progression Class 10 Exercise 1.1 Solutions Question 4.<\/strong>
    \nWhich of the following are APs? If they form an AP, find the common difference \u2019d’ and write three more terms.
    \ni) 2, 4, 8, 16, …….
    \nii) \\(2, \\frac{5}{2}, 3, \\frac{7}{2}, \\ldots \\ldots\\)
    \niii) -1.2, -3.2, -5.2, -7.2
    \niv) -10, -6, -2, 2, …….
    \nv) \\(3,3+\\sqrt{2}, 3+2 \\sqrt{2}, 3+3 \\sqrt{2} \\ldots \\ldots\\)
    \nvi) 0.2, 0.22, 0.222, 0.2222, ………
    \nvii) 0, -4, -8, -12, ………
    \nviii) \\(-\\frac{1}{2},-\\frac{1}{2},-\\frac{1}{2},-\\frac{1}{2}, \\ldots \\ldots\\)
    \nix) 1, 3, 9, 27, ……..
    \nx) a, 2a, 3a, 4a, ………..
    \nxi) a, a2<\/sup>, a3<\/sup>, a4<\/sup>, ………..
    \nxii) \\(\\sqrt{2}, \\sqrt{8}, \\sqrt{18}, \\sqrt{32} \\ldots\\)
    \nxiii) \\(\\sqrt{3}, \\sqrt{6}, \\sqrt{9}, \\sqrt{12}\\)
    \nxiv) 11<\/sup>, 32<\/sup>, 52<\/sup>, 72<\/sup>,
    \nxv) 11<\/sup>, 52<\/sup>, 72<\/sup>, 73, ……….
    \nSolution:
    \ni) 2, 4, 8, 16…………
    \nd = a2<\/sub> – a1<\/sub>= 4 – 2 = 2
    \nd = a2<\/sub> – a1<\/sub> = 8 – 4 = 4
    \nHere \u2018d\u2019 is not constant.
    \n\u2234 the given List of numbers does not form an AP.<\/p>\n

    ii) \\(2, \\frac{5}{2}, 3, \\frac{7}{2}, \\ldots \\ldots\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{2}-\\mathrm{a}_{1}=\\frac{5}{2}-\\frac{2}{1}=\\frac{5-4}{2}=\\frac{1}{2}\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{3}-\\mathrm{a}_{2}=\\frac{3}{1}-\\frac{5}{2}=\\frac{6-5}{2}=\\frac{1}{2}\\)
    \nHere \u2018d\u2019 is constant.
    \n\u2234 the given List of numbers form an AP.
    \nNext three terms are: \\(4,\\frac{9}{2}, 5\\)<\/p>\n

    iii) -1.2, -3.2. -5.2, -7.2
    \nd = a2<\/sub> – a1<\/sub> = -3.2 – (-1.2) = -3.2+ 1.2
    \nd = -2
    \nd= a3<\/sub> – a2<\/sub> = -5.2 – (-3.2) = -5.2 + 3.2
    \nd = -2
    \nHere d is constant.
    \n\u2234 This is an Arithmetic Progression.
    \nNext three terms are:
    \n-7.2 – 2 = -9.2
    \n-9.2 – 2 = -11.2
    \n-11.2 – 2 = -13.2
    \n-11.2-2 = -13.2<\/p>\n

    iv) -10. -6. -2, 2,..,…
    \nd = a2<\/sub> – a1<\/sub> = -6 – (- 10) = -6 + 10
    \nd = 4
    \nd = a3<\/sub> – a2<\/sub>= -2 – (-6) = -2 + 6
    \nd = 4
    \nHere d is constant.
    \n\u2234 Given set of numbers form an A.P.
    \nFurther three terms are
    \n2 + 4 = 6
    \n4 + 4 = 8
    \n8 + 4 = 12<\/p>\n

    v) \\(3,3+\\sqrt{2}, 3+2 \\sqrt{2}, 3+3 \\sqrt{2} \\ldots \\ldots\\)
    \nd = a2<\/sub> – a1<\/sub> = \\(=3+\\sqrt{2}-3\\)
    \n\\(\\mathrm{d}=\\sqrt{2}\\)
    \nd = a3<\/sub> – a2<\/sub> = \\(=3+2 \\sqrt{2}-(3+\\sqrt{2})\\)
    \n\\(=3+2 \\sqrt{2}-3+\\sqrt{2}\\)
    \n\\(\\mathrm{d}=\\sqrt{2}\\)
    \nHere ‘d’ is constant
    \n\u2234 Given set of numbers forms an A.P.
    \nFurther three terms are:
    \n\\(3+3 \\sqrt{2}+\\sqrt{2}=3+4 \\sqrt{2}\\)
    \n\\(3+4 \\sqrt{2}+\\sqrt{2}=3+5 \\sqrt{2}\\)
    \n\\(3+5 \\sqrt{2}+\\sqrt{2}=3+6 \\sqrt{2}\\)<\/p>\n

    vi) 0.2. 0.22. 0.222. 0.2222
    \nd = a2<\/sub> – a1<\/sub> = 0.22 – 0.2
    \nd = 0.02
    \nd = a3<\/sub> – a2<\/sub> = 0.222 – 0.22
    \nd = 0.002
    \nHere d is not constant,
    \n\u2234 Given set of numbers do not form an A.P.<\/p>\n

    vii) 0,-4, -8, -12, …………
    \na1<\/sub> = 0, a2<\/sub> = -4
    \nd = a2<\/sub> – a1<\/sub> = -4 – 0
    \nd = -4
    \nd = a3<\/sub> – a2<\/sub> = -8 – (-4) = -8 + 4
    \nd = -4
    \nHere d is constant.
    \n\u2234 Given set of numbers form an AP.
    \nFurther three terms are:
    \n-12 – 4 = -16
    \n-16 – 4 = -20
    \n-20 – 4 = -24
    \n\u2234 Next three terms are: -16. -20, -24.<\/p>\n

    viii) \\(-\\frac{1}{2},-\\frac{1}{2},-\\frac{1}{2},-\\frac{1}{2}, \\ldots \\ldots\\)
    \n\\(a_{1}=-\\frac{1}{2}, a_{2}=-\\frac{1}{2}\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{2}-\\mathrm{a}_{1}=-\\frac{1}{2}-\\left(-\\frac{1}{2}\\right)\\)
    \n\\(d=-\\frac{1}{2}+\\frac{1}{2}=0\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{3}-\\mathrm{a}_{2}=-\\frac{1}{2}-\\left(-\\frac{1}{2}\\right)\\)
    \n\\(\\mathrm{d}=-\\frac{1}{2}+\\frac{1}{2}=0\\)
    \nHere d is constant.
    \n\u2234 Given set of numbers form an AP.
    \nNext three terms are:
    \n\\(-\\frac{1}{2}+0=-\\frac{1}{2}\\)
    \n\\(-\\frac{1}{2}+0=-\\frac{1}{2}\\)
    \n\\(-\\frac{1}{2}+0=-\\frac{1}{2}\\)
    \n\u2234 Next three terms are: \\(-\\frac{1}{2},-\\frac{1}{2},-\\frac{1}{2}\\)<\/p>\n

    ix) 1, 3, 9, 27
    \nd = a2<\/sub> – a1<\/sub> = 3\u20141
    \nd = 2
    \nd = a3<\/sub> – a2<\/sub>
    \nd = 6
    \nHere \u2018d’ not is constant
    \n\u2234 Given set of numbers do not form an A.P.<\/p>\n

    x) a, 2a, 3a,4a
    \na1<\/sub>= a a2<\/sub> = 2a
    \nd = a2<\/sub> – a1<\/sub> = 2a-a
    \nd = a
    \nd = a3<\/sub> – a2<\/sub> = 3a-2a
    \nd = a
    \nHere d\u2019 is constant.
    \n\u2234 Given set of numbers form an A.P.
    \nFurther three terms are :
    \n4a + a = 5a
    \n5a + a = 6a
    \n6a + a = 7a
    \n\u2234 Next three terms are: 5a, 6a, 7a.<\/p>\n

    xi) a, a2<\/sup>, a3<\/sup>, a4<\/sup>
    \na1<\/sub> = a. a2<\/sub> = a2<\/sup>
    \nd = a2<\/sub> – a1<\/sub> = a2<\/sup> – a
    \nd= a(a – 1)
    \nd = a3<\/sub> – a2<\/sub> = a3<\/sup> – a2<\/sup>
    \nd = a2<\/sup> (a – 1)
    \nHere \u2018d is not constant.
    \n\u2234 Given set of numbers do not form an A.P.<\/p>\n

    xii) \\(\\sqrt{2}, \\sqrt{8}, \\sqrt{18}, \\sqrt{32} \\ldots\\)
    \n\\(a_{1}=\\sqrt{2}, a_{2}=\\sqrt{8}\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{2}-\\mathrm{a}_{1}=\\sqrt{8}-\\sqrt{2}\\)
    \n\\(\\mathrm{d}=2 \\sqrt{2}-\\sqrt{2}=\\sqrt{2}\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{3}-\\mathrm{a}_{2}=\\sqrt{18}-\\sqrt{8}\\)
    \n\\(\\mathrm{d}=3 \\sqrt{2}-2 \\sqrt{2}=\\sqrt{2}\\)
    \nHere d\u2019 is constant.
    \n\u2234 Given set of numbers form an A.P.
    \nSucceding three terms are :
    \n\\(\\sqrt{32}+\\sqrt{2}=4 \\sqrt{2}+\\sqrt{2}=5 \\sqrt{2}\\)
    \n\\(5 \\sqrt{2}+\\sqrt{2}=6 \\sqrt{2}\\)
    \n\\(6 \\sqrt{2}+\\sqrt{2}=7 \\sqrt{2}\\)
    \n\\(5 \\sqrt{2}, 6 \\sqrt{2}, 7 \\sqrt{2}\\)
    \n\u2234 Next three terms are: \\(\\sqrt{50}, \\sqrt{72}, \\sqrt{98}\\)<\/p>\n

    xiii) \\(\\sqrt{3}, \\sqrt{6}, \\sqrt{9}, \\sqrt{12}\\)
    \n\\(a_{1}=\\sqrt{3}, a_{2}=\\sqrt{6}\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{2}-\\mathrm{a}_{1}=\\sqrt{6}-\\sqrt{3}\\)
    \n\\(\\mathrm{d}=\\mathrm{a}_{3}-\\mathrm{a}_{2}=\\sqrt{9}-\\sqrt{6}\\)
    \n\\(\\mathrm{d}=3-\\sqrt{6}\\)
    \nHere \u2018d is not constant.
    \n\u2234 Given set of numbers do not form an A.P.<\/p>\n

    xiv) 11<\/sup>,32<\/sup>,52<\/sup>, 72<\/sup>, …………
    \n1, 9, 25, 49, ………….
    \na1<\/sub> = 1, a2<\/sub> = 9
    \nd = a2<\/sub> – a1<\/sub> = 9 – 1
    \nd = 8
    \nd = a3<\/sub> – a2<\/sub> = 25 – 9
    \nd = 16
    \nHere \u2018d\u2019 is not constant,
    \n\u2234 Given set of numbers do not form an A.P.<\/p>\n

    xv) 11<\/sup>, 52<\/sup>, 72<\/sup>, 73, ……….
    \n1, 25. 49, 73, ……….
    \na1<\/sub> = 1. a2<\/sub> = 25, a3<\/sub> = 49, a4<\/sub> = 73
    \nd = a2<\/sub> – a1<\/sub> = 25 – 1
    \nd = 24
    \nd = a3<\/sub> – a2<\/sub> = 49 – 25
    \nd = 24
    \nd = a4<\/sub> – a3<\/sub> = 73 – 49
    \nd = 24
    \nHere \u2018d\u2019 is constant.
    \n\u2234 Given set of numbers form an A.P.
    \nSucceeding three terms are:
    \n73 + 24 = 97
    \n97 + 24 = 121
    \n121 + 24 = 145
    \n\u2234 Next three terms are: 97, 121, 145.<\/p>\n

    We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

    KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.1\u00a0are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1. Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.1 KSEEB Solutions For Class 10 Maths Question …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/121"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=121"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/121\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=121"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=121"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=121"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}