{"id":26973,"date":"2021-06-28T19:02:53","date_gmt":"2021-06-28T13:32:53","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26973"},"modified":"2021-06-28T19:02:53","modified_gmt":"2021-06-28T13:32:53","slug":"ncert-solutions-for-class-10-maths-chapter-5-ex-5-4","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-5-ex-5-4\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 5 Arithmetic Progressions Ex 5.4 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nWhich term of the AP: 121, 117. 113, ….., is its first negative term?
\nSolution:
\nGiven sequence is 121,117,113,…
\nIt is an AP. with a = 121, d = 117 – 121 = – 4
\nLet nth term of this sequence be its first negative term.
\nSo, a < 0
\ni.e., a + (n – 1 )d < 0
\n121 + (n – 1) – 4 < 0
\n121 – 4n + 4 < 0
\n125 < 4n \u21d2 4n > 125
\n\u21d2 n > \\(\\frac { 125 }{ 4 }\\) = 31.25
\nHence, 32th<\/sup> term of the given sequence is its first negative term.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nThe sum of the third term and the seventh term of an AP is 6 and their product is 8. Fibd the sum of first sixteen terms of the AP?
\nSolution:
\nThird term of AP is a + (3 – 1 )d = a + 2d
\nSeventh term of AP is a + (7 – 1)d = a + 6d
\nWe have,
\nSum of third and seventh terms of AP
\n(a + 2d) + (a + 6 d) = 6
\n2a + 8d= 6
\na + 4d= 3
\n\u21d2 a = 3 – 4d
\nAlso given,
\nProduct (a + 2d) (a + 6d)= 6
\na\u00b2 + 6ad + 2ad + 12d\u00b2 = 8
\na\u00b2 + 8ad + 12d\u00b2 = 8
\nNow put the value of a in equation, we get,
\n(3 – 4d)\u00b2 + 8(3 – 4d) + 12d\u00b2 = 8
\n\u21d2 a + 16d\u00b2 – 24d + 24d – 32d\u00b2 + 12d\u00b2 = 8
\na – 4d\u00b2 = 8
\n\u21d2 4d\u00b2 = 1
\n\u21d2 d\u00b2 = \\(\\frac { 1 }{ 4 }\\)
\nSo, d = \u00b1\\(\\frac { 1 }{ 2 }\\)
\nTaking d = \u00b1\\(\\frac { 1 }{ 2 }\\), Put the value of d in
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nA ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 at the top. If the top and the bottom rungs are \\({ 2 }\\frac { 1 }{ 2 } m\\) apart, what is the length of the wood required for the rungs? [Hint: Number of rungs = \\(\\frac { 250 }{ 25 }\\)]
\n\"NCERT
\nSolution:
\nLength of the rungs of a ladder decreases uniformly from bottom to top.
\nDistance between bottom to top
\n= 2\\(\\frac { 1 }{ 2 }\\)m
\n= \\(\\frac { 5 }{ 2 }\\) x 100cm = 250cm
\nHence, total no. of rungs = \\(\\frac { 250 }{ 25 }\\) = 10
\nSeries is 45,45 + d, 45 + 2d,… 45 + 9d
\nFirst term = a = 45, last term = 45 + 9d
\nAccording to question,
\n45 + 9d = 25
\n45 – 25 = 9d
\n20 = 9d
\nd = \\(\\frac { -20 }{ 9 }\\)
\nTherefore, length of the wood required for rungs is,
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nThe houses of a row are numbered consecutively from 1 to 49. Show that there is value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
\nSolution:
\nThe series of houses will be 1, 2, 3, 4, …… 49
\nSx-1<\/sub> = S49<\/sub> – Sx<\/sub>
\nLet x be the number
\nAccording to the given condition
\n\"NCERT
\nNow, put the values of Sx-1<\/sub>, S49<\/sub> and (Sx-1<\/sub> = S49<\/sub> – Sx<\/sub>)
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nA small terrace at a football ground comprises of 15 steps each of which is 50\u00a0 m long and built of solid concrete. Each step has a rise of \\(\\frac { 1 }{ 2 } m\\) and a tread of \\(\\frac { 1 }{ 2 } m\\). Calculate the total volume of concrete required to build the terrace.
\nSolution:
\nVolume of concrete required to build the first step.
\n= \\(\\frac { 1 }{ 4 }\\) x \\(\\frac { 1 }{ 2 }\\) x 50m\u00b3
\nVolume of concrete required to build the second step
\n= \\(\\frac { 2 }{ 4 }\\) x \\(\\frac { 1 }{ 2 }\\) x 50m\u00b3
\nSimilarly, third step = \\(\\frac { 3 }{ 4 }\\) x \\(\\frac { 1 }{ 2 }\\) x 50m\u00b3
\nand for 15th = \\(\\frac { 15 }{ 4 }\\) x \\(\\frac { 1 }{ 2 }\\) x 50m\u00b3
\nSo, total volume of concrete required to build 15 steps
\n\"NCERT
\nHence, the total volume of concrete required to build the terrace is 750 m\u00b3.<\/p>\n

\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Exercise 5.4 Question 1. Which term of the AP: 121, 117. 113, ….., is its first negative term? Solution: Given sequence …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-5-ex-5-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4 Questions and Answers are prepared by our highly skilled subject experts. 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