{"id":10701,"date":"2024-02-28T15:25:39","date_gmt":"2024-02-28T09:55:39","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=10701"},"modified":"2024-02-29T17:54:35","modified_gmt":"2024-02-29T12:24:35","slug":"ts-10th-class-maths-solutions-chapter-6-progressions-ex-6-4","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-solutions-chapter-6-progressions-ex-6-4\/","title":{"rendered":"TS 10th Class Maths Solutions Chapter 6 Progressions Ex 6.4"},"content":{"rendered":"

Students can practice TS Class 10 Maths Solutions<\/a> Chapter 6 Progressions Ex 6.4 to get the best methods of solving problems.<\/p>\n

TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.4<\/h2>\n

Question 1.
\nIn which of the following situations, does the list of numbers involved in form a G.P. ?<\/p>\n

i) Salary of Sharmila, when her salary is Rs. 5,00,000 for the first year and expected to receive yearly increase of 10%.
\nSolution:
\nGiven : Sharmila’s yearly salary = Rs. 5,00,000
\nRate of annual increment = 10%
\n\"TS
\nHere, a = a1<\/sub> = 5,00,000
\n\"TS
\nEvery term starting from the second can be obtained by multiplying its preceding term by a fixed number \\(\\frac{11}{10}\\).
\n\u2234 r = common ratio = \\(\\frac{11}{10}\\)
\nHence the situation forms a G.R<\/p>\n

ii) Number of bricks needed to make each step, if the stair case has total 30 steps. Bottom step needs 100 bricks and each successive step needs 2 bricks less than the previous step.
\nSolution:
\nGiven : Bricks needed for the bottom step = 100.
\nEach successive step needs 2 bricks less than the previous step.
\n\u2234 Second step from the bottom needs = 100 – 2 = 98 bricks.
\nThird step from the bottom needs = 98 – 2 = 96 bricks
\nFourth step from the bottom needs = 96 – 2 = 94 bricks.
\nHere the numbers are
\n100, 98, 96, 94, ………….
\nClearly this is an A.P. but not G.R<\/p>\n

\"TS<\/p>\n

iii) Perimeter of the each triangle, when the mid-points of sides of an equilateral triangle whose side is 24 cm are joined to form another triangle, whose mid-points in turn are joined to form still another triangle and the process continues indefinitely.
\n\"TS
\nSolution:
\nPerimeter of the 1st<\/sup> equilateral triangle = 3 \u00d7 24 = 72 cm
\nPerimeter of the 2nd<\/sup> equilateral triangle = 3 \u00d7 12 = 36 cm
\nPerimeter of the 3rd<\/sup> equilateral triangle = 3 \u00d7 6 = 18 cm
\nSuccessive terms are obtained by dividing with 2 the preceding term except first term.
\n\u2234 The above situation is a G.P.<\/p>\n

Question 2.
\nWrite three terms of the G.P. when the first term ‘a’ and the common ratio ‘r’ are given.<\/p>\n

i) a = 4; r = 3.
\nSolution:
\nThe terms are a, ar, ar2<\/sup>, ar3<\/sup>.
\n\u2234 4, 4 \u00d7 3, 4 \u00d7 32<\/sup>, 4 \u00d7 33<\/sup>
\n\u2234 4, 12, 36, 108,…………<\/p>\n

ii) a = \\(\\sqrt{5}\\); r = \\(\\frac{1}{5}\\)
\nSolution:
\nThe terms of a G.P are:
\na, ar, ar2<\/sup>, ar3<\/sup>,……….
\n\"TS<\/p>\n

iii) a = 81; r = \\(-\\frac{1}{3}\\).
\nSolution:
\nThe terms of a G.P are:
\na, ar, ar2<\/sup>,…………
\n\u21d2 81, 81 \u00d7 \\(\\left(\\frac{-1}{3}\\right)\\), 81 \u00d7 \\(\\left(\\frac{-1}{3}\\right)^2\\)
\n\u21d2 81, -27, 9,…………<\/p>\n

iv) a = \\(\\frac{1}{64}\\), r = 2.
\nSolution:
\nGiven: a = \\(\\frac{1}{64}\\) ; r = 2
\na1<\/sub> = a = \\(\\frac{1}{64}\\)
\na2<\/sub> = ar = \\(\\frac{1}{64}\\) \u00d7 2 = \\(\\frac{1}{32}\\)
\na3<\/sub> = ar2<\/sup> = \\(\\frac{1}{64}\\) \u00d7 22<\/sup> = \\(\\frac{1}{64}\\) \u00d7 4 = \\(\\frac{1}{16}\\)
\n\u2234 The G.P. is \\(\\frac{1}{16}\\), \\(\\frac{1}{32}\\), \\(\\frac{1}{16}\\),………….<\/p>\n

Question 3.
\nWhich of the following are G.P.? If they are G.P, write three more terms.<\/p>\n

i) 4, 8, 16,………….
\nSolution:
\nGiven : 4, 8, 16,………….
\nwhere, a1<\/sub> = 4; a2<\/sub> = 8; a3<\/sub> = 16,………….
\n\\(\\frac{a_2}{a_1}\\) = \\(\\frac{8}{4}\\) = 2
\n\\(\\frac{\\mathrm{a}_3}{\\mathrm{a}_2}\\) = \\(\\frac{16}{8}\\) = 2
\n\u2234 r = \\(\\frac{a_2}{a_1}\\) = \\(\\frac{a_3}{a_2}\\) = 2
\nHence 4,8,16,……… is a G.P
\nwhere a = 4 and r = 2
\na4<\/sup> = a. r3<\/sup> = 4 \u00d7 23<\/sup> = 4 \u00d7 8 = 32
\na5<\/sup> = a.r4<\/sup> = 4 \u00d7 24<\/sup> = 4 \u00d7 16 = 64
\na6<\/sup> = a. r5<\/sup> = 4 \u00d7 25<\/sup> = 4 \u00d7 32 = 128<\/p>\n

ii) \\(\\frac{1}{3}\\), \\(\\frac{-1}{6}\\), \\(\\frac{1}{12}\\)………, (x \u2260 0)
\nSolution:
\nGiven: t1<\/sub>= \\(\\frac{1}{3}\\); t2<\/sub> = \\(\\frac{-1}{6}\\): t3<\/sub> = \\(\\frac{1}{12}\\),………….
\n\"TS
\nHence the ratio is common between any two successive terms.
\n\u2234 \\(\\frac{1}{3}\\), \\(\\frac{-1}{6}\\), \\(\\frac{1}{12}\\),………. is a G.P.
\n\"TS<\/p>\n

iii) 5, 55, 555,………….
\nSolution:
\nGiven: t1<\/sub> = 5, t2<\/sub> = 55, t3<\/sub> = 555
\n\\(\\frac{\\mathrm{t}_2}{\\mathrm{t}_1}\\) = \\(\\frac{55}{5}\\) = 11
\n\\(\\frac{\\mathrm{t}_3}{\\mathrm{t}_2}\\) = \\(\\frac{555}{55}\\) = \\(\\frac{111}{11}\\)
\n\u2234 \\(\\frac{t_2}{t_1}\\) \u2260 \\(\\frac{t_3}{t_2}\\)<\/p>\n

iv) -2, -6, -18,…………..
\nSolution:
\nGiven : t1<\/sub> = -2; t2<\/sub> = -6; t3<\/sub> = -18
\n\\(\\frac{\\mathrm{t}_2}{\\mathrm{t}_1}\\) = \\(\\frac{-6}{-2}\\) = 3; \\(\\frac{t_3}{t_2}\\) = \\(\\frac{-18}{-6}\\) = 3
\n\u2234 r = \\(\\frac{t_2}{t_1}\\) = \\(\\frac{t_3}{t_2}\\) = …….. = 3
\n\u2234 -2, -6, -18,……….. is a G.P.
\nwhere a = -2 and r = 3
\nan<\/sub> =a.rn-1<\/sup>
\na4<\/sub> = ar3<\/sup> = (-2) \u00d7 33<\/sup> = -2 \u00d7 27 = -54
\na5<\/sub> = ar4<\/sup> = (-2) \u00d7 34<\/sup> = – 2 \u00d7 81 = -162
\na6<\/sub> = a.r5<\/sup> = (-2) \u00d7 35<\/sup> = -2 \u00d7 243 = -486<\/p>\n

v) \\(\\frac{1}{2}\\), \\(\\frac{1}{4}\\), \\(\\frac{1}{6}\\), …………..
\nSolution:
\nGiven: t1<\/sub> = \\(\\frac{1}{2}\\), t2<\/sub> = \\(\\frac{1}{4}\\), t3<\/sub> = \\(\\frac{1}{6}\\)
\n\\(\\frac{t_2}{t_1}\\) = \\(\\frac{\\frac{1}{4}}{\\frac{1}{2}}\\) = \\(\\frac{1}{4}\\) \u00d7 \\(\\frac{2}{1}\\) = \\(\\frac{1}{2}\\)
\n\\(\\frac{t_3}{t_2}\\) = \\(\\frac{\\frac{1}{6}}{\\frac{1}{4}}\\) = \\(\\frac{1}{6}\\) \u00d7 \\(\\frac{4}{1}\\) = \\(\\frac{2}{3}\\)
\n\u2234 \\(\\frac{\\mathrm{t}_2}{\\mathrm{t}_1}\\) \u2260 \\(\\frac{t_3}{t_2}\\)
\ni.e., \\(\\frac{1}{2}\\), \\(\\frac{1}{4}\\), \\(\\frac{1}{6}\\),………… is not a G.P.<\/p>\n

\"TS<\/p>\n

vi) 3, -32<\/sup>, 33<\/sup>,…………
\nSolution:
\nGiven : t1<\/sub> = 3; t2<\/sub> = -32<\/sup>, t3<\/sub> = 33<\/sup>,….
\n\\(\\frac{t_2}{t_1}\\) = \\(\\frac{-3^2}{3}\\) = -3
\n\\(\\frac{\\mathrm{t}_3}{\\mathrm{t}_2}\\) = \\(\\frac{3^3}{-3^2}\\) = -3
\n\u2234 r = \\(\\frac{t_2}{t_1}\\) = \\(\\frac{t_3}{t_2}\\) = …….. = -3
\ni.e., every term is obtained by multiplying its preceding term by a fixed number -3.
\n3, -32<\/sup>, 33<\/sup>, ……… forms a G.P
\nwhere a = 3; r = -3
\nan<\/sup> = a.rn-1<\/sup>
\n\u2234 a4<\/sub> = 3 \u00d7 (-3)4-1<\/sup> = 3 \u00d7 (-3)3<\/sup> = -81
\na5<\/sub> = 3 \u00d7 (-3)4<\/sup> = 3 \u00d7 81 = 243
\na6<\/sub> = 3 \u00d7 (-3)5<\/sup> = 3 \u00d7 (-243) = -729<\/p>\n

vii) x, 1, \\(\\frac{1}{x}\\),……… (x \u2260 0)
\nSolution:
\nGiven:
\n\"TS
\n\"TS<\/p>\n

viii) \\(\\frac{1}{\\sqrt{2}}\\), -2, \\(\\frac{8}{\\sqrt{2}}\\),………..
\nSolution:
\nGiven :
\n\"TS
\nGiven terms are not in G.P.<\/p>\n

ix) 0.4, 0.04, 0.004,………
\nSolution:
\nGiven : t1<\/sub> = 0.4; t2<\/sub> = 0.04; t3<\/sub> = 0.004,…………..
\n\"TS
\n\"TS<\/p>\n

Question 4.
\nFind x so that x, x + 2, x + 6 are consecutive terms of a geometric progression.
\nSolution:
\nGiven x, x + 2 and x + 6 are in G.P but read it as x, x + 2 and x + 6.
\n\u2234 r = \\(\\frac{t_2}{t_1}\\) = \\(\\frac{t_3}{t_2}\\)
\n\u21d2 \\(\\frac{x+2}{x}\\) = \\(\\frac{x+3}{x+2}\\)
\n\u21d2 (x + 2)2<\/sup> = x(x + 6)
\n\u21d2 x2<\/sup> + 4x + 4 = x2<\/sup> + 6x
\n\u21d2 4x – 6x = -4 \u21d2 -2x = -4
\n\u2234 x = 2<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can practice TS Class 10 Maths Solutions Chapter 6 Progressions Ex 6.4 to get the best methods of solving problems. TS 10th Class Maths Solutions Chapter 6 Progressions Exercise 6.4 Question 1. In which of the following situations, does the list of numbers involved in form a G.P. ? i) Salary of Sharmila, when … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10701"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=10701"}],"version-history":[{"count":5,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10701\/revisions"}],"predecessor-version":[{"id":10951,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10701\/revisions\/10951"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=10701"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=10701"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=10701"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}