{"id":12348,"date":"2022-06-06T01:00:38","date_gmt":"2022-06-05T19:30:38","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=12348"},"modified":"2022-05-23T16:11:49","modified_gmt":"2022-05-23T10:41:49","slug":"arithmetic-progressions-class-10-extra-questions","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/arithmetic-progressions-class-10-extra-questions\/","title":{"rendered":"Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Solutions Answers"},"content":{"rendered":"

Here you will find Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 10 Maths<\/a> are solved by experts and will guide students in the right direction.<\/p>\n

Extra Questions for Class 10 Maths Arithmetic Progressions with Answers Solutions<\/h2>\n

Extra Questions for Class 10 Maths Chapter 5 Arithmetic Progressions with Solutions Answers<\/strong><\/p>\n

Arithmetic Progressions Class 10 Extra Questions Objective Type<\/h3>\n

Question 1.
\nCommon difference for arithmetic progression (A.P) 3, 1, -1, -3 ….. will be:
\n(a) 1
\n(b) 2
\n(c) -2
\n(d) 3
\nAnswer:
\n(c) -2
\nSolution.
\nCommon difference = a2<\/sub> – a1<\/sub>
\n= 1 – 3 = -2
\nHence, choice (iii) is correct<\/p>\n

\"Arithmetic<\/p>\n

Question 2.
\nCommon difference of 2, 0.5, -1, -2.5, -4,….. is:
\n(a) -1.5
\n(b) -1.3
\n(c) 2.4
\n(d) 2.6
\nAnswer:
\n(a) -1.5<\/p>\n

Question 3.
\n100th term of \\(\\frac {1}{2}\\), 1, \\(\\frac {3}{2}\\), 2, ….is:
\n(a) 50
\n(b) 60
\n(c) 70
\n(d) 40
\nAnswer:
\n(a) 50<\/p>\n

Question 4.
\nrth<\/sup> term of a + 2b, a – b, a – 46, … is:
\n(a) a + (5 – 3r)b
\n(b) a + (4 – 3r)b
\n(c) a + (6 – r)b
\n(d) a + (2 – r)b
\nAnswer:
\n(a) a + (5 – 3r)b<\/p>\n

Question 5.
\nThe sum of first 16 terms of the AP 10, 6, 2, …. is:
\n(a) 320
\n(b) -320
\n(c) -352
\n(d) -400
\nAnswer:
\n(b) -320<\/p>\n

\"Arithmetic<\/p>\n

Question 6.
\nThe sum of first 20 terms of the AP 1, 3, 5, 7, 9, … is
\n(a) 264
\n(b) 400
\n(c) 472
\n(d) 563
\nAnswer:
\n(b) 400<\/p>\n

Question 7.
\n(5 + 13 + 21 + … + 181) =?
\n(a) 2476
\n(b) 2337
\n(c) 2219
\n(d) 2139
\nAnswer:
\n(d) 2139<\/p>\n

Question 8.
\nThe sum of n terms of the AP \u221a2, \u221a8, \u221a18, \u221a32, … is
\n(a) 1
\n(b) 2n(n + 1)
\n(c) \\(\\frac {1}{2}\\)n(n + 1)
\n(d) \\(\\frac{1}{\\sqrt{2}}\\)n(n + 1).
\nAnswer:
\n(d) \\(\\frac{1}{\\sqrt{2}}\\)n(n + 1)<\/p>\n

Question 9.
\nHow many terms of the AP 3, 7, 11, 15, ….will make the sum 406?
\n(a) 10
\n(b) 12
\n(c) 14
\n(d) 20
\nAnswer:
\n(c) 14<\/p>\n

\"Arithmetic<\/p>\n

Question 10.
\nThe sum of first 100 natural numbers is
\n(a) 4950
\n(b) 5050
\n(c) 5000
\n(d) 5150
\nAnswer:
\n(b) 5050<\/p>\n

Question 11.
\nThe sum of all odd numbers between 100 and 200 is
\n(a) 3750
\n(b) 6200
\n(c) 6500
\n(d) 7500
\nAnswer:
\n(d) 7500<\/p>\n

Question 12.
\nThe sum of all even natural numbers less than 100 is
\n(a) 2272
\n(b) 2352
\n(c) 2450
\n(d) 2468
\nAnswer:
\n(c) 2450<\/p>\n

Question 13.
\nThe sum of first fifteen multiples of 8 is
\n(a) 840
\n(b) 960
\n(c) 872
\n(d) 1080
\nAnswer:
\n(b) 960<\/p>\n

\"Arithmetic<\/p>\n

Question 14.
\nIn an AP, the first term is 22, nth term is 11 and the sum of first n terms is 66. The value of n is
\n(a) 10
\n(b) 12
\n(c) 14
\n(d) 16
\nAnswer:
\n(b) 12<\/p>\n

Question 15.
\nIn an AP, the first term is 8, nth term is 33 and the sum of first n terms is 123. Then, d =?
\n(a) 5
\n(b) – 5
\n(c) 7
\n(d) 3
\nAnswer:
\n(a) 5<\/p>\n

Question 16.
\nThe sum of n terms of an AP is given by Sn<\/sub> = (2n2<\/sup> + 3n). What is the common difference of the AP?
\n(a) 3
\n(b) 4
\n(c) 5
\n(d) 9
\nAnswer:
\n(b) 4<\/p>\n

Arithmetic Progressions Class 10 Extra Questions Very Short answer Type<\/h3>\n

Question 1.
\nFor the following APs, write the first term and the common difference.
\n(i) 3, 1, -1, -3, …
\n(ii) – 5, -1, 3, 7, …..,
\n(iii) \\(\\frac{1}{3}, \\frac{5}{3}, \\frac{9}{3}, \\frac{13}{3}\\) ,…..
\n(iv) 0.6, 1.7, 2.8, 3.9, …
\nSolution.
\n(i) First term a = t1<\/sub> = 3, Common difference
\nd = 2nd term – 1st term
\n= 1 – 3 = -2<\/p>\n

(ii) First term a = t1<\/sub> = -5, Common difference
\nd = 2nd term – 1st term
\n= -1 -(-5) = 1 + 5 = 4<\/p>\n

(iii) First term a = t1<\/sub> = \\(\\frac {1}{3}\\) Common difference
\nd = 2nd term – 1st term = \\(\\frac{5}{3}-\\frac{1}{3}=\\frac{4}{3}\\)<\/p>\n

(iv) First term a = t1<\/sub> = 0.6 Common difference
\nd = 2nd term – 1st term
\n= 1.7 – 0.6 = 1.1<\/p>\n

\"Arithmetic<\/p>\n

Question 2.
\nFor what value of k will k + 9, 2k – 1, and 2k + 7 are the consecutive terms of an arithmetic progression (A.P)
\nSolution.
\nGiven, k + 9, 2k – 1 and 2k + 7 are the consecutive term
\n\u2234 d = a2<\/sub> – a1<\/sub> = a3<\/sub> – a2<\/sub>
\n\u21d2 (2k – 1) – (k + 9) = (2k + 7) – (2k – 1)
\n\u21d2 2k – 1 – k – 9 = 2k + 7 – 2k + 1
\nk – 10 = 8
\n\u2234 k = 8 + 10 = 18<\/p>\n

Question 3.
\nFor what value of k, will the points (k, -1), (2, 1) and (4, 5) lie on a line.
\nSolution.
\nIf point (k, -1), (2, 1) and (4, 5) lie on a line, i.e., the points are collinear.
\nx1<\/sub> = k, y1<\/sub> = -1, x2<\/sub> = 2, y2<\/sub> = 1, x3<\/sub> = 4, y3<\/sub> = 5
\nThe area of A formed by these points are zero.
\n\u2234 0 = \\(\\frac {1}{2}\\) [x1<\/sub> (y2<\/sub> – y3<\/sub>) + x2<\/sub>(y3<\/sub> – y1<\/sub>) + x3<\/sub>(y1<\/sub> – y2<\/sub>)]
\n\u21d2 k(1 -5) + 2(5 + 1) + 4 (-1 -1) = 0
\n\u21d2 -4k + 12 – 8 = 0
\n\u21d2 -4k = -4
\n\u21d2 k = 1<\/p>\n

\"Arithmetic<\/p>\n

Question 4.
\nFind the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
\nSolution.
\nlet a be the first term and d the common difference of an AP. Now, the nth term of an AP is
\nan<\/sub> = a + (n – 1)d
\na11<\/sub> = a + 10d = 38
\n[\u2235 a11<\/sub> = 38 (given)] …(i)
\nand a16<\/sub> = a + 15d = 73
\n[\u2235 a16<\/sub> = 73 (given)] ..(ii)
\nOn subtracting Eq. (i) from EQuestion (ii) we get
\n5d = 35 = d = \\(\\frac {35}{5}\\) = 7
\nFrom Eq. (i)
\na + 10 \u00d7 7 = 38
\n\u21d2 a = 38 – 70 = – 32
\n\u2234 The 31st term of an AP
\na31<\/sub> = a + 30d
\n= -32 + 30 \u00d7 7
\n= – 32 + 210 = 178<\/p>\n

\"Arithmetic<\/p>\n

Question 5.
\nFind the sum of the following AP’s
\n(i) 2, 7, 12, …, to 10 terms
\n(ii) -37, -33, -29, …, to 12 terms
\n(iii) 0.6, 1.7, 2.8, … to 100 terms
\n(iv) \\(\\frac{1}{15}, \\frac{1}{12}, \\frac{1}{10}\\) ……., to 11 terms.
\nSolution.
\n(i) Let a be the first term and d be the common difference of the given AP.
\nThen, we have a = 2 and d = 7 – 2 = 5
\n\u2235 Sum of n terms of AP,
\nSn<\/sub> = \\(\\frac {n}{2}\\)[2a + (n – 1)d]
\nPutting a = 2, d = 5, n = 10, we get
\nS10<\/sub> = \\(\\frac {10}{2}\\)[2 \u00d7 2 + (10 – 1)5]
\n= 5 (4 + 9 \u00d7 5)
\n= 5 (4 + 45) = 5 \u00d7 49 = 245
\n(ii) let a be the first and d be the common difference of the given AP,
\nThen, we have a = -37,
\nd = -33 -(-37) = – 33 + 37
\n= 4
\n\u2235 Sum of n terms of an AP,
\nSn<\/sub> = \\(\\frac {n}{2}\\)[2a + (n – 1)d], we get
\nS12<\/sub> = [2 \u00d7 (-37) + (12 – 1)4]
\n= 6(-74 + 11 \u00d7 4)
\n= 6 – 74 +44)
\n= 6 \u00d7 (-30) = – 180.
\n(iii) Let a be the first term and d be the common difference of the given AP.
\nThen, we have a = 0.6, d = 1.7 -0.6 = 1.1
\n\u2235 Sum of n terms of an AP,
\nSn<\/sub> = \\(\\frac {n}{2}\\)[2a + (n – 1)d], we get
\nS100<\/sub> = \\(\\frac {100}{2}\\) [2 \u00d7 0.6 + (100 – 1) 1.1]
\n= 50 (1.2 + 99 \u00d7 1.1)
\n= 50 (1.2 + 108.9)
\n= 50 \u00d7 110.1
\n= 5505
\n(iv) Let a be the first term and d be the common difference of the given AP.
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-1\"
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-2\"<\/p>\n

\"Arithmetic<\/p>\n

Question 6.
\nThe first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
\nSolution.
\nLet a be the first term and d be the common difference of an AP.
\nGive, first term a = 5, last term l = 45 and sum of n terms Sn<\/sub> = 400
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-3\"
\n\u2234 The number of terms is 16 and the common difference is \\(\\frac {8}{3}\\).<\/p>\n

Question 7.
\nThe first and the last terms of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there and what is their sum?
\nSolution.
\nLet a be the first term and d be the common difference.
\nGiven, first term a = 17, last term l = an<\/sub> = 350, common difference d = 9
\n\u2235 l = an<\/sub> = 350
\n\u21d2 a + (n – 1)d = 350
\n[\u2235 l = an<\/sub> = a + (n – 1)d]
\n\u21d2 17 + (n – 1)9 = 350 =
\n\u21d2 9(n – 1) = 350 – 17 = 333
\n\u21d2 n – 1 = \\(\\frac {333}{9}\\) = 37
\n\u21d2 n = 37 + 1 = 38
\nPutting a = 17,1 = 350, n = 38
\n\u2235 Sum of n terms
\nSn<\/sub> = \\(\\frac {n}{2}\\)(a + l), we get
\nS39<\/sub> = \\(\\frac {38}{2}\\) (17 + 350)
\n= 19 (367) = 6973
\nSo, there are 38 terms in the AP having their sum as 6973.<\/p>\n

\"Arithmetic<\/p>\n

Question 8.
\nWhich term of the AP 3, 8, 13, 18, …. is 78?
\nSolution.
\nLet nth term, be 78. Given, 3, 8, 13, 18 …. are in AP.
\nFirst term a, = 3, common difference, d = 8 – 3 = 5
\n\u2235 nth term an<\/sub> = 78
\n\u2234 a + (n – 1) )d = 78
\n\u21d2 3 + (n – 1) 5 = 78
\n\u21d2 (n – 1) 5 = 78 – 3 5
\n\u21d2 (n – 1) = 75 n – 1 = 15
\n\u21d2 n = 15 + 1
\n\u21d2 n = 16
\nHence, 16th term be 78.<\/p>\n

Question 9.
\nFind the number of terms in each of the following APs
\n(i) 7, 13, 19, …., 205
\n(ii) 18, 15 \\(\\frac {1}{2}\\), 13, …., -47
\nSolution.
\n(i) Suppose, there are n terms in the given AP. Then nth term an<\/sub> = 205, first term a = 7, common difference d = 13 – 7 = 6.
\n\u2235 a + (n – 1) d = an<\/sub>
\n\u2234 a + (n – 1)d = 205
\n\u21d2 7 + (n – 1)6 = 205
\n\u21d2 6(n – 1) = 205 – 7
\n\u21d2 6(n – 1) = 198
\n\u21d2 n – 1 = \\(\\frac {198}{6}\\) = 33
\nn = 33 + 1 = 34
\n\u21d2 Hence, the given AP contains 34 terms.<\/p>\n

(ii) Suppose, there are n terms in the given AP
\nGiven, first term a = 18, common difference
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-4\"
\nThen nth term an<\/sub> = -47
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-5\"
\n= -13 \u00d7 -2 = 26
\n\u21d2 n = 26 + 1 = 27
\nHence, the given AP contains 27 terms.<\/p>\n

Arithmetic Progressions Class 10 Extra Questions Short answer Type<\/h3>\n

Question 1.
\nFind the sum of all odd integers between z and 100 which are divisible by 3.
\nSolution.
\nThe odd numbers which are divisible by 3, between 2 and 100 are 3, 9, 15, 21, 27, 33,…..99 which clearly form on A.P. here a = 3, d = 6 and l = 99
\n\u2234 99 = a + (n – 1)d
\n\u21d2 99 = 3 + (n – 1)d
\n\u21d2 (n – 1)6 = 99 – 3 = 96
\n\u21d2 n = \\(\\frac {96}{6}\\) + 1 = 17
\nNow, to find the sum of 17 terms
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-6\"
\n= 17 \u00d7 51
\n= 867.<\/p>\n

\"Arithmetic<\/p>\n

Question 2.
\nWhich term of arithmetic progression 21, 18, 15…. is – 81? Is zero an<\/sub>y term of this arithmetic progress?
\nSolution.
\nGiven AP be 21, 18, 15…… Let nth<\/sup> term of this progression is – 81
\na = 21, d = -3
\n\u2234 -81 = 21 + (n – 1) \u00d7 -3
\n\u21d2 -81 – 21 = -3(n – 1)
\n\u21d2 n – 1 = \\(\\frac {102}{3}\\) = 34
\n\u2234 n = 34 + 1
\n= 35
\nHence 35th<\/sup> term of this progression is 81
\nLet Tn<\/sub> = 0 = 21 + (n – 1)(-3)
\n\u21d2 21 – 3n + 3 = 0
\n\u2234 3n = 24
\nn = 8
\nthen a 8th<\/sup> term of the progression is 0.<\/p>\n

Question 3.
\nHow many terms should taken from the series 9, 17, 25, 33… to get a total of 636?
\nSolution.
\nThe given series an<\/sub> A.P. where a = 9, d 25 – 17 = 8 and Sn<\/sub> = 636.
\n\u21d2 Sn<\/sub> \\(\\frac {n}{2}\\)[2a + (n – 1)d]
\n\u21d2 636 = \\(\\frac {n}{2}\\)[2 \u00d7 9 + (n – 1) 8]
\n\u21d2 636 = n [9 + 4n – 4]
\n\u21d2 4n2<\/sup> + 5n – 636 = 0
\n\u21d2 4n2<\/sup> + (53 – 48) n – 636 = 0
\n\u21d2 4n2<\/sup> + 53n – 48n – 636 = 0
\n\u21d2 n(4n + 53)- 12 (4n + 53) = 0
\n\u21d2 (4n + 53) (n – 12) = 0
\n\u2234 n = 12 or – \\(\\frac {53}{4}\\)
\nas number of terms are not infraction and – ve
\n\u2234 n = 12
\nHence sum of 12 term of the series = 636.<\/p>\n

\"Arithmetic<\/p>\n

Question 4.
\nA contract on construction job specifies a penalty for delay of completion beyond a certain date as follows \u20b9 200 for the first day, \u20b9 250 for the second day, \u20b9 300 for the third day etc., the penalty for each succeeding day being \u20b9 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
\nSolution.
\nSince, the penalty for each succeeding day is 50 more than the preceding day, therefore the penalties for the first day, the second day, the third day, etc. will form an AP.
\nLet us denote the penalty for the nth day by an<\/sub>, then
\na1<\/sub> = 200, a2<\/sub> = \u20b9 250, a3<\/sub> = \u20b9 300
\nHere, a = \u20b9 200, d = \u20b9 250 – \u20b9 200 = \u20b9 50 and n = 30.
\n\u2234 The money the contractor has to pay penalty, if he delayed the work by 30 days.
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-7\"
\n= 15 (400 + 29 \u00d7 50)
\n= 15 (400 + 1450) = 15 \u00d7 1850.
\n= 27750
\nSo, a delay of 30 days costs is \u20b9 27750<\/p>\n

Question 5.
\nA sum of \u20b9 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is \u20b9 20 less than its preceding prize, find the value of each of the prizes.
\nSolution.
\nSuppose, the respective prizes are a + 60, a + 40, a + 20, a, a – 20, a – 40, a – 60
\nAccording to question,
\na + 60 + a + 40 + a + 20 + a + a – 20 + a – 40 + a – 60 = 700
\n\u21d2 7a = 700
\n\u21d2 a = \\(\\frac {700}{7}\\) = 100
\nHence, the seven prizes are 100 + 60, 100 + 40, 100 + 20, 100, 100 – 20, 100 – 40, 100 – 60 .’i.e., 160, 140, 120, 100, 80, 60, 40 .<\/p>\n

\"Arithmetic<\/p>\n

Question 6.
\nTwo APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
\nSolution.
\nLet the Two APs a1<\/sub>, a2<\/sub>, a3<\/sub>…….an<\/sub> and b1<\/sub>, b2<\/sub>, b3<\/sub>, …. bn<\/sub>
\nAlso, let d be the same common difference of two APs. then
\nthe nth term of first AP
\nOn = a1<\/sub> + (n-1)d and
\nthe nth term of second AP
\nbn<\/sub> = b1<\/sub> + (n – 1)d
\nNow, an<\/sub> – bn<\/sub> = [a1<\/sub> + (n – 1)d] – [b1<\/sub> + (n – 1)d]
\n\u21d2 an<\/sub> – bn<\/sub> = a1<\/sub> – by for all n \u2208 N
\n\u21d2 a100<\/sub> – b100<\/sub> = a1<\/sub> – b1<\/sub> = 100 (Given)
\n\u2234 a1000<\/sub> – b1000<\/sub> = a1<\/sub> – b1<\/sub>
\n\u21d2 a1000<\/sub> – b1000<\/sub> = 100 [\u2235 a1<\/sub> – b1<\/sub> = 100]
\nHence, the difference between their 1000th terms is also 100 for all n \u2208 N.<\/p>\n

\"Arithmetic<\/p>\n

Question 7.
\nHave many three digit numbers are divisible by 7?
\nSolution.
\nWe know that, 105 is the first and 994 is the last 3 digit number divisible by 7. Thus, we have to determine the number of terms in the list 105, 112, 119, …, 994.
\nClearly, the successive difference of terms is constant with common difference
\nd = 112 – 105 = 7
\nSo, it forms an AP.
\nLet there be n terms, in the AP, then nth term = 994
\n\u2235 an<\/sub> = a + (n – 1)d
\n\u21d2 105 + (n – 1)7 = 994
\n\u21d2 7(n – 1) = 994 – 105
\n\u21d2 7(n – 1) = 889
\n\u21d2 n – 1 = \\(\\frac {889}{7}\\)
\n= 127
\n\u21d2 n = 127 + 1 = 128
\nSo, there are 128 numbers of three digit which are divisible by 7.<\/p>\n

\"Arithmetic<\/p>\n

Question 8.
\nHow many multiple of 4 lie between 10 and 250?
\nSolution.
\nWe see that 12 is the first integer between 10 and 250, which is a multiple of 4. Also, when we divide 250 by 4, the remainder is 2. Therefore, 250 – 2 = 248 is the greatest integer divisible by 4 and lying between 10 and 250. Thus, we have to find the number of terms in an AP whose first term = 12, last term = 248 and common difference = 4.
\nLet the n term of the AP, is
\nan<\/sub> = 248
\n= 12 + (n – 1) 4 = 248
\n[\u2235 an<\/sub>= a + (n – 1)d]
\n\u21d2 4(n – 1) = 248 – 12
\n\u21d2 4(n – 1) = 236
\n\u21d2 n – 1 = \\(\\frac {236}{4}\\) = 59
\n\u21d2 n = 59 + 1 = 60
\nHence, there are 60 multiples of 4 lie between 10 and 250.<\/p>\n

Arithmetic Progressions Class 10 Extra Questions Long answer Type<\/h3>\n

Question 1.
\nA spiral is made up of successive semicircles with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, …. as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semi-circles?
\n[Take, \u03c0 = \\(\\frac {22}{7}\\) ]
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-8\"
\nSolution.
\nLength of spiral made up of thirteen consecutive semi-circles
\n= (\u03c0 \u00d7 0.5 + \u03c0 \u00d7 1.0 + \u03c0 \u00d7 1.5 + \u03c0 \u00d7 2.0 + … + \u03c0 \u00d7 6.5)
\n= \u03c0 \u00d7 0.5 (1 + 2 + 3 + … + 13)
\nwhich form an AP with first term, a = 1,
\ncommon difference, d = 2 – 1 = 1
\nand number of term, n = 13
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-9\"
\n= 143 cm.<\/p>\n

\"Arithmetic<\/p>\n

Question 2.
\n200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-10\"
\nSolution.
\nSince, logs are stacked in each row form a series 20 + 19 + 18 + 17 + …. Clearly, it is an AP with first term, a = 20 and common difference, d = 19 – 20 = -1
\nSuppose, Sn<\/sub> = 200
\n\u2235 Sn<\/sub> = \\(\\frac {n}{2}\\)[2a + (n – 1) d]
\n\u21d2 200 = \\(\\frac {n}{2}\\)[2 \u00d7 20 + (n – 1) (-1)]
\n\u21d2 400 = n(40 – n + 1)
\n\u21d2 n2<\/sup> – 41n + 400 = 0
\n\u21d2 n2<\/sup> – 25n – 16n + 400 = 0
\n(By factorization method)
\n\u21d2 n(n-25) – 16 (n – 25) = 0
\n\u21d2 (n – 25) (n – 16) = 0
\n\u21d2 n = 16
\nor n = 25
\nHence, the number of rows is either 25 or 16.
\nWhen n = 16, tn<\/sub> = a + (n – 1)
\n= 20 + (16 – 1) (-1)
\n= 20 – 15
\n= 5
\nWhen n = 25,
\ntn<\/sub> = a+ (n – l) d
\n= 20 + (25 – 1) (-1)
\n20 – 24 = -4 (Not possible)
\nHence, the number of row is 16 and number of logs in the top row = 5.<\/p>\n

\"Arithmetic<\/p>\n

Question 3.
\nIn a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in lines (see figure) A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
\n\"Arithmetic-Progressions-Class-10-Extra-Questions-Maths-Chapter-5-with-Solutions-Answers-11\"
\nSolution.
\nAccording to question, a competitor pick up the Ist potato, second potato, third potato, fourth potato ….
\nThe distances sum by competitor are 2 \u00d7 5, 2 \u00d7 (5 + 3), 2 \u00d7 (5 + 3 + 3), 2 \u00d7 (5 + 3 + 3 + 3) i.e., 10, 16, 22, 28, …..
\nClearly, it is an AP with first term, a = 10
\nand common difference, d = 16 – 10 = 6
\n\u2235 The sum of n terms,
\nSn<\/sub> = \\(\\frac {n}{2}\\)[2a + (n – 1) d]
\n\u2234 The sum of 10 terms,
\nS10<\/sub> = \\(\\frac {10}{2}\\)[2 \u00d7 10 + (10 – 1) \u00d7 6]
\n[\u2235 n = 10 (given)]
\n= 5 (20 + 54)
\n= 5 \u00d7 74
\n= 370
\nHence, the total distance the competitor has to run = 370 m.<\/p>\n

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

Here you will find Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. Extra Questions for Class 10 Maths Arithmetic Progressions with Answers Solutions Extra Questions for Class 10 Maths Chapter 5 Arithmetic Progressions …<\/p>\n

Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Solutions Answers<\/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":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nArithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Solutions Answers - MCQ Questions<\/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:\/\/mcq-questions.com\/arithmetic-progressions-class-10-extra-questions\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Solutions Answers - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"Here you will find Arithmetic Progressions Class 10 Extra Questions Maths Chapter 5 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction. 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