{"id":6044,"date":"2023-08-12T05:04:40","date_gmt":"2023-08-12T05:04:40","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6044"},"modified":"2023-08-12T05:04:40","modified_gmt":"2023-08-12T05:04:40","slug":"ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-4\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 5 Arithmetic Progressions Ex 5.4"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 5 (Arithmetic Progressions)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 5 Arithmetic Progressions <\/strong>Exercise 5.4 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 5 Arithmetic Progressions<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 5.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 5.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-5-arithmetic-progressions-ex-5-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 5.3<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 5.4<\/span><\/strong><\/h2>\n<p>&nbsp;<\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>1. Which term of the AP: 121, 117, 113, . . ., is its first negative term?<br \/>\n[Hint: Find n for a<sub>n<\/sub>\u00a0&lt; 0]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\nthe AP series is 121, 117, 113, . . .,<br \/>\n<em>a<\/em> = 121<br \/>\n<em>d<\/em> = 117 &#8211; 121= -4<br \/>\n<em>a<sub>n<\/sub><\/em>\u00a0=\u00a0<em>a <\/em>+ (<em>n<\/em>\u00a0\u22121)<em>d<br \/>\n<\/em>a<sub>n<\/sub> = 121 + (n\u22121)(-4)<br \/>\na<sub>n<\/sub> = 121 &#8211; 4n + 4<br \/>\na<sub>n<\/sub> = 125 &#8211; 4n<br \/>\nTo find the first negative term of the series,\u00a0<em>a<sub>n\u00a0<\/sub><\/em>&lt; 0<br \/>\n125 &#8211; 4n &lt; 0<br \/>\n125 &lt; 4n<br \/>\nn &gt; 125\/4<br \/>\nn &gt; 31.25<br \/>\nTherefore, the first negative term of the series is the 32<sup>nd<\/sup>\u00a0term.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>2. The sum of the third and the seventh terms of an AP is 6, and their product is 8. Find the sum of the first sixteen terms of the AP.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211; <\/strong><em>n<sup>th<\/sup>\u00a0<\/em>term formula, <em>a<sub>n<\/sub> = a + (n \u2212 1)d<br \/>\nThird term, a<sub>3<\/sub> = a + (3 &#8211; 1)d<br \/>\na<sub>3<\/sub>\u00a0= a + 2d<br \/>\n<\/em>Seventh term<em>, a<sub>7 <\/sub>= a + <\/em>(7 &#8211; 1)<em>d<br \/>\na<sub>7 <\/sub>= a + 6d<br \/>\n<\/em>According to given conditions,<br \/>\na<sub>3<\/sub>\u00a0+ a<sub>7<\/sub> = 6\u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (i)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a<sub>3<\/sub>\u00a0\u00d7 a<sub>7<\/sub> = 8\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (ii)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Substituting the values in eqn. (i), we get,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a + 2d + a + 6d = 6<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">2a + 8d = 6<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a + 4d = 3<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = 3 \u2013 4d\u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (iii)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now substituting the values in eqn. (ii), we get,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(a + 2d) \u00d7 (a + 6d) = 8\u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; (iv)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Putting the value of a from eqn. (iii) in eqn. (iv), we get,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(3 \u2013 4d + 2d) \u00d7 (3 \u2013 4d + 6d) = 8<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">(3 \u2013 2d) \u00d7 (3 + 2d) = 8<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">3<sup>2<\/sup>\u00a0\u2013 (2d)<sup>2<\/sup>\u00a0= 8 \u00a0 \u00a0 \u00a0 \u00a0 (using the identity,\u00a0<em>(a + b)(a &#8211; b)\u00a0=\u00a0a<sup>2 <\/sup>&#8211; b<sup>2<\/sup><\/em>)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">9 \u2013 4d<sup>2<\/sup>\u00a0= 8<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">4d<sup>2<\/sup> = 9 &#8211; 8 = 1<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">d = \u221a(1\/4)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">d = \u00b11\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">d = 1\/2 or -1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">So now, if<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">d = 1\/2, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = 3 \u2013 4d <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = 3 \u2013 4(1\/2) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = 3 \u2013 2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = 1<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">and if d = -1\/2,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = 3 \u2013 4d <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = 3 \u2013 4(-1\/2) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = 3 + 2 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">a = 5<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Sum of n<sup>th<\/sup>\u00a0term of AP is:\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\na = 1 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">d = 1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Then, the sum of first 16 terms are;<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>16<\/sub> <\/em>= 16\/2 [2 + (16-1)1\/2] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>16<\/sub> <\/em>= 8(2 + 15\/2)\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">S<sub>16<\/sub>\u00a0= 76<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">And\u00a0when <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>a<\/em> = 5 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>d <\/em>= -1\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Then, the sum of first 16 terms are;<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>n<\/sub><\/em>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><br \/>\nS<sub>16<\/sub> = 16\/2 [2.5 + (16-1)(-1\/2)] <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">S<sub>16<\/sub> = 8(5\/2)\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">S<sub>16<\/sub>\u00a0= 20<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>3. A ladder has rungs 25 cm apart (see Fig. 5.7). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{2\\frac{1}{2}}\" alt=\"\\mathbf{2\\frac{1}{2}}\" align=\"absmiddle\" \/>m<\/strong><strong>\u00a0apart, what is the length of the wood required for the rungs?<br \/>\n[Hint: Number of rungs = -250\/25 ].<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6300\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.4-Que3.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"233\" height=\"310\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.4-Que3.png 233w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.4-Que3-225x300.png 225w\" sizes=\"auto, (max-width: 233px) 100vw, 233px\" \/><\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong>Given:<br \/>\nDistance between the rungs of the ladder is 25 cm.<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Distance between the top rung and bottom rung of the ladder will be (in cm) = 2 \u00bd \u00d7 100 = 250 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, the total number of rungs = 250\/25 + 1 = 11 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">As we can observe here, that, the ladder has rungs in decreasing order from top to bottom. Thus, the rungs are decreasing in an order of AP. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">So, According to given condition\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>a<\/em> = 45<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>l<\/em> = 25<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Number of terms, <em>n<\/em> = 11<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now, as we know, sum of n<sup>th<\/sup>\u00a0terms is equal to,\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>n<\/sub><\/em> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;a&amp;space;+&amp;space;l)&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ a + l) \\right ]\" align=\"absmiddle\" \/><br \/>\n<em>S<sub>n<\/sub><\/em>= 11\/2(45 + 25) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>n <\/sub><\/em>= 11\/2(70) <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>n <\/sub><\/em>= 385 cm<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence,\u00a0the length of the wood required for the rungs is 385cm.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a 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.<br \/>\n[Hint :S<sub>x \u2013 1<\/sub> = S<sub>49<\/sub> \u2013 S<sub>x<\/sub>]<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong>Given :<br \/>\nRow houses are numbers from 1, 2, 3, 4, 5 \u2026\u2026. 49.<br \/>\nThus, we can see the houses numbered in a row are in the form of AP.<br \/>\n<em>a<\/em> = 1<br \/>\n<em>d <\/em>= 1<br \/>\nLet us say the number of the houses can be represented as<br \/>\nSum of n<sup>th<\/sup> term of AP.<br \/>\n<em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><\/em><br \/>\nSum of number of houses beyond x house = <em>S<sub>x &#8211; 1<br \/>\n<\/sub><\/em><em>S<sub>x &#8211; 1 <\/sub><\/em>= (x &#8211; 1)\/2[2(1) + (x &#8211; 1 &#8211; 1)1]<br \/>\n<em>S<sub>x &#8211; 1 <\/sub><\/em>= (x &#8211; 1)\/2 [2 + x &#8211; 2]<br \/>\n<em>S<sub>x &#8211; 1 <\/sub><\/em>= x(x &#8211; 1)\/2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (i)<br \/>\nBy the given condition, we can write<br \/>\n<em>S<sub>49\u00a0<\/sub>\u2013 S<sub>x<\/sub>\u00a0<\/em>= {49\/2[2(1) + (49 &#8211; 1)1]} \u2013 {x\/2[2(1) + (x &#8211; 1)1]}<br \/>\n<em>S<sub>49\u00a0<\/sub>\u2013 S<sub>x<\/sub>\u00a0<\/em>= 25(49) \u2013 x(x + 1)\/2\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212; (ii)<br \/>\nAs per the given condition, eq.(i) and eq(ii) are equal to each other.<br \/>\nTherefore,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6302\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.4-Ans4.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"189\" height=\"141\" \/><br \/>\nAs we know, the number of houses cannot be a negative number. Hence, the value of x is 35.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>5. A small terrace at a football ground comprises of 15 steps, each of which is 50 m long and built of solid concrete. Each step has a rise of 1 4 m and a tread of 1 2 m (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = \u00bc \u00d71\/2 \u00d750 m<sup>3<\/sup>.]<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-6301\" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.4-Que5.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"449\" height=\"153\" srcset=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.4-Que5.png 449w, https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex5.4-Que5-300x102.png 300w\" sizes=\"auto, (max-width: 449px) 100vw, 449px\" \/><br \/>\n<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><strong>Solution &#8211;\u00a0 <\/strong>As we can see from the given figure,\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">1<sup>st<\/sup>\u00a0step is \u00bc m high, \u00bd m wide and 50 m long<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">2<sup>nd<\/sup>\u00a0step is (\u00bc+\u00bc = \u00bd m) high, \u00bd m wide and 50 m long and,\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">3<sup>rd<\/sup>\u00a0step is (3\u00d7\u00bc = 3\/4 m) high, \u00bd m wide and 50 m long.\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, we can conclude that the height of step increases by \u00bc m each time when width and length is \u00bd m and 50 m respectively.\u00a0 <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">So, the height of steps forms a series AP in such a way that; <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">\u00bc, \u00bd , \u00be, 1, 5\/4, \u2026\u2026.. <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Volume of steps = Volume of Cuboids = Length \u00d7 Breadth \u00d7 Height<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now, <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Volume of concrete required to build the first step = \u00bc \u00d7\u00bd \u00d750 = 25\/4<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Volume of concrete required to build the second step =\u00bd \u00d7\u00bd\u00d750 = 25\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Volume of concrete required to build the second step = \u00be \u00d7 \u00bd \u00d750 = 75\/2<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Volume of steps = \u00bd \u00d7 50 \u00d7 (\u00bc + 2\/4 + 3\/4 + 4\/4 + 5\/4 + \u2026\u2026.)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;- (i)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Now, we can see the height of concrete required to build the steps, are in AP series;<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Thus, applying the AP series concept to the height,<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>a<\/em> = 1\/4<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>d<\/em> = 2\/4-1\/4 = 1\/4<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>n<\/em> = 15<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">As, the sum of n terms is <\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>n<\/sub>\u00a0= <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{n}{2}\\left&amp;space;[&amp;space;2a+(n-1)d&amp;space;\\right&amp;space;]\" alt=\"\\frac{n}{2}\\left [ 2a+(n-1)d \\right ]\" align=\"absmiddle\" \/><\/em><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>15 <\/sub><\/em>= 15\/2(2\u00d7(1\/4 ) + (15 &#8211; 1)1\/4)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\"><em>S<sub>15 <\/sub><\/em>= 15\/2 (4)<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">S<sub>n<\/sub>\u00a0= 30<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Hence, up solving eqn. (1), we get<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">Volume of steps = \u00bd \u00d7 50 \u00d7 S<sub>n<br \/>\n<\/sub>= \u00bd \u00d7 50 \u00d7 30<\/span><br \/>\n<span style=\"color: #000000; font-family: Georgia, Palatino;\">= 750 m<sup>3<br \/>\n<\/sup>Hence,\u00a0the total volume of concrete required to build the terrace is 750 m<sup>3<\/sup>.<\/span><\/p>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/a><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 5 (Arithmetic Progressions)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 5 Arithmetic Progressions Exercise 5.4 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 5 Arithmetic Progressions NCERT Class 10 Maths Solution Ex &#8211; 5.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1074,1078,1044,1049,1048],"class_list":["post-6044","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-5-arithmetic-progressions-solutions","tag-ncert-class-10-mathematics-exercise-5-4-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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