{"id":22308,"date":"2020-02-04T12:43:10","date_gmt":"2020-02-04T07:13:10","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=22308"},"modified":"2021-07-02T15:25:51","modified_gmt":"2021-07-02T09:55:51","slug":"1st-puc-basic-maths-question-bank-chapter-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-question-bank-chapter-5\/","title":{"rendered":"1st PUC Basic Maths Question Bank Chapter 5 Progressions"},"content":{"rendered":"

Students can Download Basic Maths Chapter 5 Progressions Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Basic Maths Question Bank Chapter 5 Progressions<\/h2>\n

Arithmetic Progression (A.P)<\/span><\/p>\n

Question 1.
\nWhich term of A.P 4\\(\\left(5 \\frac{1}{3}-4\\right), 6 \\frac{2}{3} \\ldots \\ldots\\) is 104?
\nAnswer:
\nIn given A.P: a = 4, d = 51\/3<\/sup> – 4 = \\(\\frac { 16 }{ 3 }\\) – 4 = \\(\\frac { 4 }{ 3 }\\)
\nLet nth<\/sup> term be 104 = Tn<\/sub> = a + (n – 1) d
\n\u21d2 4 + (n – 1) \\(\\frac { 4 }{ 3 }\\) = 104
\n\u21d2 12 + 4n – 4 = 312 \u21d2 4n – 4 = 312
\n\u21d2 4n = 312 – 8
\n\u21d2 4n = 304
\n\u2234 n = 76<\/p>\n

Question 2.
\nHow many terms of series 24 + 20 + 16 + ……. must be taken so that their sum is 72.
\nAnswer:
\nIn the given A.P; a = 24, d = 4, Sn<\/sub> = 72
\nwe have Sn<\/sub> = \\(\\frac { n }{ 2 }\\){2a + (n – 1)d}
\n72 = \\(\\frac { n }{ 2 }\\){2(24) + (n – 1)(-4)}
\n= 144 = n{48 – 4n + 4}
\n\u21d2 144 = n{52 – 4n}
\n\u21d2 144 = 52n – 4n2
\n<\/sup>\u21d2 4n2<\/sup> – 52n + 144 = 0
\n\u21d2 n2<\/sup> – 13n + 36 = 0
\n\u21d2 (n – 9)(n – 4) = 0
\n\u2234 n = 9,4<\/p>\n

\"1st<\/p>\n

Question 3.
\nThe first and the last term of an A.P are -4 and 146 and the sum of the A.P is 7171. Find the number of terms in the A.P and the A.P.
\nAnswer:
\nGiven a = -4, l = 146.
\nWe have l = a + (n – 1 )d = 146 = -4 + (n – 1) d
\n\u21d2 (n – 1)d = 150
\nAlso we have Sn = \\(\\frac { n }{ 2 }\\){2a + (n – 1)d}
\n7171 = \\(\\frac { n }{ 2 }\\){2(-4) + 150}
\n14342 = n{-8 + 150}
\n\u21d2 142n = 14342
\n\u2234 n = \\(\\frac{14342}{142}\\) = 101
\nSo number of terms is A.P = 101
\nconsider (n – 1)d = 150.
\n\"1st<\/p>\n

Question 4.
\nFind the sum of all numbers between 100 and 1000 which are divisible by 11.
\nAnswer:
\nThe 1st<\/sup> greatest No. greater than 100 is 110. the last which is less than 1000 is, 990.
\n110,121, ………990 be the number
\nTn<\/sub> = 90 \u21d2 110 + (n – 1) 11 = 990
\n11n – 11 = 990 – 110
\n11n = 880 + 11 = 891
\n\u2234 n = \\(\\frac { 891 }{ 11 }\\) = 81
\nNow S81<\/sub> = 8\\(\\frac { 1 }{ 2 }\\)(110 + 990) = \\(\\frac { 81 }{ 2 }\\)(550) = 4450<\/p>\n

Question 5.
\nThe first, second and the last terms of an A.P are a, b, c respectively prove that the sum is \\(\\frac{(a+c)(b+c-2 a)}{2(b-a)}\\)
\nAnswer:
\n1st<\/sup> term is a, common difference = d = b – a.
\nLet ‘n’ be the number of term in the series.
\n\u2234 c = a + (n – 1)d
\n\u21d2 (n – 1)d = c – a
\n\u21d2 (n – 1 )(b – a) = c – a
\n\u21d2 n – 1 = \\(\\frac{c-a}{b-a}\\)
\n\u21d2 n = \\(\\frac{c-a}{b-a}\\) + 1
\n= \\(\\frac{n}{2}\\){2a + c – a} = \\(\\frac{n}{2}\\){a + c} = \\(\\left(\\frac{b+c-2 a}{b-a}\\right)\\left(\\frac{a+c}{2}\\right)\\)<\/p>\n

Geometric Progression (G.P)<\/span><\/p>\n

Question 1.
\nFind the 9th<\/sup> term and sum of 6 term of the geometric sequence \\(\\frac{1}{3}, \\frac{1}{9}, \\frac{1}{27} \\ldots \\ldots\\)
\nAnswer:
\na = \\(\\frac{1}{3}\\) r = \\(\\frac{1}{3}\\)
\n\"1st<\/p>\n

\"1st<\/p>\n

Question 2.
\nThe second and fifth term of GP are 3 and \\(\\frac{81}{8}\\) respectively. Find the GP
\nAnswer:
\nGiven T2<\/sub> = ar = 3, T5<\/sub> = ar4<\/sup> = \\(\\frac{81}{8}\\)
\n\"\"
\n\u2234 r = \\(\\sqrt[3]{\\frac{33}{23}}=\\frac{3}{2}\\)
\nIf ar = 3 \u21d2 a. \\(\\frac{3}{2}\\) = 3 \u21d2 a = 2
\nG.p is 2, 3, \\(\\frac{9}{2}\\), …….<\/p>\n

Question 3.
\nThe sum of three numbers which are in G.P is \\(\\frac{39}{10}\\) and their product is 1. Find the numbers.
\nAnswers:
\nLet the number be \\(\\frac{a}{2}\\), a, ar. which are in G.P.
\ni.e., \\(\\frac{a}{2}\\) + a + ar = \\(\\frac{39}{10}\\)
\n\"1st
\n\u21d2 a3<\/sup> = 1
\n\u2234 a = 1
\nput a = 1 in (1),
\n\\(\\frac{1}{r}\\) + 1 + r = \\(\\frac{39}{10}\\)
\n\u21d2 \\(\\frac{1+r+r^{2}}{r}=\\frac{39}{10}\\)
\n\u21d2 10 + 10r + 10r3 <\/sup>= 39
\nr \u21d2 10r2<\/sup> – 29r + 10 = 0
\n\u21d2 r = \\(\\frac{5}{2}\\) and \\(\\frac{2}{5}\\)
\n\"1st<\/p>\n

\"1st<\/p>\n

Question 4.
\nFind three numbers in GP. Whose sum is 35 and sum of whose squares is 525.
\nAnswer:
\nLet the three numbers be a, ar, ar2<\/sup> which are is GP.
\nGiven a + ar + ar2<\/sup> = 35 \u21d2 a( 1 + r + r2<\/sup>) = 35
\nand a2<\/sup> + a2<\/sup>r2<\/sup> + a2<\/sup>r4<\/sup> = 525
\n\u21d2 a2<\/sup>( 1 + r2<\/sup> + r4<\/sup>) = 525
\n\\(\\frac{a^{2}\\left(1+r^{2}+r^{4}\\right)}{a\\left(1+r+r^{2}\\right)}=\\frac{525}{35}=15\\)
\na(1 + r2<\/sup> + r4<\/sup>) = 15(1 + r + r2<\/sup>)
\n\"1st
\n\u21d2 a(1 – r + r2<\/sup>) = 15 …….. (2)
\nAdd (1) and (2)
\n2a + 2ar2<\/sup> = 50
\nall (2) – (1) \u21d2 2ar = -20
\n\"1st
\n\\(\\frac{1}{r}\\) + r = \\(\\frac{-5}{2}\\)
\n\u21d2 2(1 + r2<\/sup>) = -5r
\n\u21d2 2r2<\/sup> + 5r + 2 = 0
\n\u21d2 r = \\(\\frac{-5 \\pm \\sqrt{25-4(2)(2)}}{4}\\)
\n\u21d2 \\(\\frac{-5 \\pm 3}{4}=\\frac{-1}{2},-2\\)
\nwhen r = -2 2ar = -20 \u21d2 -4a = -20 [\u2234 a = 5]
\nwhen r = \\(-\\frac{1}{2}\\) , 2ar = -20 \u21d2 -a = -20 \u21d2 [a = 20]
\nHence the three number which are in G.P are 5, -10, 20 (or) 20, – 10, 5<\/p>\n

Question 5.
\nFind the sum of n terms of 5 + 55 + 555 + ……..
\nAnswer:
\nLet s = 5 + 55 + 555 + …….. n terms
\n\u21d2 s = 5[1 + 11 + 111 + …… to term
\n= \\(\\frac{5}{9}\\)(9 + 99 + 999 + ….. to n term) \\(\\frac{5}{9}\\)[(10 – 1) + (100 – 1) + (1000 – 1) + …… to terms]
\n= \\(\\frac{5}{9}\\)[10 + 102<\/sup> + 103<\/sup> + …… to n terms] + (1 + 1 + ……. to n terms)
\n\"1st<\/p>\n

\"1st<\/p>\n

Harmonic Progression (H.P)<\/span><\/p>\n

Question 1.
\nIf the 12th<\/sup> element of H.P is \\(\\frac { 1 }{ 5 }\\) and the 19th term is \\(\\frac { 3 }{ 32 }\\). Find the 10th<\/sup> term of H.P.
\nAnswer:
\n12th<\/sup> term = \\(\\frac { 1 }{ 15 }\\)
\n19th<\/sup> term = \\(\\frac{3}{2^{2}}\\)
\n\u2234 12th<\/sup> term, AP is 5 and
\n19th<\/sup> termof A.P is \\(\\frac { 22 }{ 3 }\\)
\n\"1st<\/p>\n

Question 2.
\nIf 1, x, 3 are in H.P . Find x.
\nAnswer:
\nGiven 1, x, 3 are in H.P
\n\"1st<\/p>\n

Question 3.
\nIf the mth<\/sup> term of H.P is ‘n’ and nth<\/sup> term is m, prove that (m + n)th<\/sup> is \\(\\frac{m n}{m+n}\\)
\nAnswer:
\nGiven mth<\/sup> term of H.P is n,
\n\u2234 mth<\/sup> term of A.P = \\(\\frac{1}{n}\\) = a + (m – 1) d
\nAlso nth<\/sup> term of HP is m.
\n\u2234 nth<\/sup> term of A.P = \\(\\frac{1}{m}\\) = a + (n – 1)d
\na + (m – 1)d = \\(\\frac{1}{n}\\) …… (1)
\na + (n – 1)d = \\(\\frac{1}{m}\\) …… (2)
\nSolving (1) and (2)
\n\"1st<\/p>\n

\"1st<\/p>\n

Arithmetic, Geometric and Hormonic Means.<\/span><\/p>\n

Question 1.
\nFind six geometrical mean between 27 and \\(\\frac { 1 }{ 81 }\\)
\nAnswer:
\nLet the required GM is g1<\/sub>, g2<\/sub>, g3<\/sub>, g4<\/sub>, g5<\/sub>, g6<\/sub>
\n= 27, g1<\/sub>, g2<\/sub>, g3<\/sub>, g4<\/sub>, g5<\/sub>, g6<\/sub>, \\(\\frac{1}{81}\\) are in GP
\nNow a = 27 and \\(\\frac{1}{81}\\) = 8th <\/sup>element
\n\u21d2 \\(\\frac{1}{81}\\) = ar7<\/sup> \u21d2 \\(\\frac{1}{81}\\) = 27.r7<\/sup>
\n\u21d2 \\(\\frac{1}{81 \\times 27}\\) = r7<\/sup>
\n\u21d2 r7<\/sup> = \\(\\frac{1}{3^{7}}\\)
\n\u21d2 r = \\(\\frac{1}{3}\\)
\n\u2234 the six GM are 9, 3, 1,\\(\\frac{1}{3}, \\frac{1}{9}, \\frac{1}{27}\\)<\/p>\n

Question 2.
\nIf a, b, c are in A.P and a, mb, c are in GP then show that a, m2<\/sup>b, c are in H.P
\nAnswer:
\na, b, c, are in AP .
\n\u21d2 b = \\(\\frac{a+c}{2}\\)
\na, mb, c are in GP,
\n\u21d2 mb = \\(\\sqrt{a c}\\)
\n\u21d2 m2<\/sup>b2<\/sup> = ac
\n\u21d2 m2<\/sup>b. b. = ac
\n\u21d2 m2<\/sup>b \\(\\left(\\frac{a+c}{2}\\right)\\) = ac \\(\\left(\\because b=\\frac{a+c}{2}\\right)\\)
\n\u21d2 m2<\/sup>b = \\(\\frac{2 a c}{a+c}\\)
\n\u21d2 a, m2<\/sup>b, c are in H.P.<\/p>\n

\"1st<\/p>\n

Summation of Series<\/span><\/p>\n

Question 1.
\n12<\/sup> + 32<\/sup> + 52<\/sup> + ………\u00a0 to \u2019n’ terms.
\nAnswer:
\n1, 3, 5, ………. are in A.P
\nTn<\/sub> = a + (n – 1)d = 1 + (n – 1)2 = 1 + 2x – 2 = 2x – 1
\n\u2234 nth<\/sup> term = (2n – 1)2<\/sup>
\nTn<\/sub> = (2x – 1)2<\/sup> = 4n2<\/sup> – 4x + 1
\nrequired sum = Sn<\/sub> = \u03a3Tn<\/sub> = \u03a34n2<\/sup> – 4x + 1 = 4\u03a3n2<\/sup> – 4\u03a3n + \u03a31.
\n\\(=\\frac{4 n(n+1)(2 n+1)}{6}-\\frac{4 n(n+1)}{2}+n\\)
\n= \\(\\frac{n}{3}\\) [2(n + 1)(2n + 1) – 6(n + 1) + 3] = \\(\\frac{n}{3}\\)[4n2<\/sup> – 1]<\/p>\n

Question 2.
\nFind the sum to ‘n’ terms the series 1.2 + 4.5 + 7.8 + ….. to term.
\nAnswer:
\n1st factors are 1, 4, 7,
\n\u2234 nth<\/sup>term = 1 + (n – 1) 3 = 3n – 2
\n2nd factor are 2, 5, 8,
\n\u2234 nth<\/sup> term = 2 + (n – 1) 3 = 3n – 1
\n\u21d2 nth<\/sup> term of the series is (3n – 2)(3n – 1)
\n\u2234 The sum = \u03a3(3n – 2)(3n – 1)
\n= \u03a3(9x2<\/sup> – 9x + 2) = 9\u03a3n2<\/sup> -9\u03a3n + 2\u03a31
\n\\(=9 \\frac{n(n+1)(2 n+1)}{6}-9 \\frac{n(n+1)}{2}+2 n\\)
\n= \\(\\frac{1}{2}\\)n(n + 1)[3(2n + 1) – 9] + 2n = n(3n2<\/sup> – 1).<\/p>\n

\"1st<\/p>\n

Question 3.
\nSum the series \\(\\frac{1^{3}}{1}+\\frac{1^{3}+2^{3}}{1+3}+\\frac{1^{3}+2^{3}+3^{3}}{1+3+5}+\\ldots \\ldots .16 \\text { terms }\\) 16terms
\nAnswer:
\n\"1st<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Chapter 5 Progressions Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Basic Maths Question Bank Chapter 5 Progressions Arithmetic Progression (A.P) Question 1. Which term of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22308"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=22308"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22308\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=22308"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=22308"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=22308"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}