{"id":119146,"date":"2020-12-02T15:06:40","date_gmt":"2020-12-02T09:36:40","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=119146"},"modified":"2020-12-02T15:07:34","modified_gmt":"2020-12-02T09:37:34","slug":"rs-aggarwal-solutions-class-10-chapter-11-ex-11d","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/rs-aggarwal-solutions-class-10-chapter-11-ex-11d\/","title":{"rendered":"RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D"},"content":{"rendered":"
These Solutions are part of RS Aggarwal Solutions Class 10<\/a>. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions<\/p>\n Very-Short and Short-Answer Questions<\/strong> Question 2.<\/strong><\/span> Question 3.<\/strong><\/span> Question 4.<\/strong><\/span> Question 5.<\/strong><\/span> Question 6.<\/strong><\/span> Question 7.<\/strong><\/span> Question 8.<\/strong><\/span> Question 9.<\/strong><\/span> Question 10.<\/strong><\/span> Question 11.<\/strong><\/span> Question 12.<\/strong><\/span> Question 13.<\/strong><\/span> Question 14.<\/strong><\/span> Question 15.<\/strong><\/span> Question 16.<\/strong><\/span> Question 17.<\/strong><\/span> Question 18.<\/strong><\/span> Question 19.<\/strong><\/span> Question 20.<\/strong><\/span> Question 21.<\/strong><\/span> Question 22.<\/strong><\/span> Question 23.<\/strong><\/span> Question 24.<\/strong><\/span> Question 25.<\/strong><\/span> Question 26.<\/strong><\/span> Question 27.<\/strong><\/span> Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions<\/a>\u00a0are helpful to complete your math homework.<\/p>\n If you have any doubts, please comment below. A Plus Topper<\/a> try to provide online math tutoring for you.<\/p>\n","protected":false},"excerpt":{"rendered":" RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Exercise 11D Very-Short and Short-Answer Questions Question 1. Solution: (3y – 1), (3y + 5) and (5y+ 1) are in AP … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6805],"tags":[],"yoast_head":"\nExercise 11D<\/strong><\/span><\/h3>\n
\nQuestion 1.<\/strong><\/span>
\nSolution:<\/strong><\/span>
\n(3y – 1), (3y + 5) and (5y+ 1) are in AP
\n(3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)
\n\u21d2 2 (3y + 5) = (5y + 1) + (3y – 1)
\n\u21d2 6y + 10 = 8y
\n\u21d2 8y – 6y = 10
\n\u21d2 2y = 10
\n\u21d2 y = 5
\ny = 5<\/p>\n
\nSolution:<\/strong><\/span>
\nk, (2k – 1) and (2k + 1) are the three successive terms of an AP.
\n(2k – 1) – k = (2k + 1) – (2k – 1)
\n\u21d2 2 (2k – 1) = 2k + 1 + k
\n\u21d2 4k – 2 = 3k + 1
\n\u21d2 4k – 3k = 1 + 2
\n\u21d2 k = 3
\nk = 3<\/p>\n
\nSolution:<\/strong><\/span>
\n18, a, (b – 3) are in AP
\n\u21d2 a – 18 = b – 3 – a
\n\u21d2 a + a – b = -3 + 18
\n\u21d2 2a – b = 15<\/p>\n
\nSolution:<\/strong><\/span>
\na, 9, b, 25 are in AP.
\n9 – a = b – 9 = 25 – b
\nb – 9 = 25 – b
\n\u21d2 b + b = 22 + 9 = 34
\n\u21d2 2b = 34
\n\u21d2 b= 17
\na – b = a – 9
\n\u21d2 9 + 9 = a + b
\n\u21d2 a + b = 18
\n\u21d2 a + 17 = 18
\n\u21d2 a = 18 – 17 = 1
\na = 18, b= 17<\/p>\n
\nSolution:<\/strong><\/span>
\n(2n – 1), (3n + 2) and (6n – 1) are in AP
\n(3n + 2) – (2n – 1) = (6n – 1) – (3n + 2)
\n\u21d2 (3n + 2) + (3n + 2) = 6n – 1 + 2n – 1
\n6n + 4 = 8n – 2
\n\u21d2 8n – 6n = 4 + 2
\n\u21d2 2n = 6
\n\u21d2 n = 3
\nand numbers are
\n2 x 3 – 1 = 5
\n3 x 3 + 2 = 11
\n6 x 3 – 1 = 17
\ni.e. (5, 11, 17) are required numbers.<\/p>\n
\nSolution:<\/strong><\/span>
\nThree digit numbers are 100 to 990 and numbers which are divisible by 7 will be
\n105, 112, 119, 126, …, 994
\nHere, a = 105, d= 7, l = 994
\nTn<\/sub> = (l) = a + (n – 1) d
\n\u21d2 994 = 105 + (n – 1) x 7
\n\u21d2 994 – 105 = (n – 1) 7
\n\u21d2 (n – 1) x 7 = 889
\n\u21d2 n – 1 = 127
\n\u21d2 n = 127 + 1 = 128
\nRequired numbers are 128<\/p>\n
\nSolution:<\/strong><\/span>
\nThree digit numbers are 100 to 999
\nand numbers which are divisible by 9 will be
\n108, 117, 126, 135, …, 999
\nHere, a = 108, d= 9, l = 999
\nTn\u00a0<\/sub>(l) = a + (n – 1) d
\n\u21d2 999 = 108 + (n – 1) x 9
\n\u21d2 (n – 1) x 9 = 999 – 108 = 891
\n\u21d2 n – 1 = 99
\n\u21d2 n = 99 + 1 = 100<\/p>\n
\nSolution:<\/strong><\/span>
\nSum of first m terms of an AP = 2m\u00b2 + 3m
\nSm<\/sub>\u00a0= 2m\u00b2 + 3m
\nS1<\/sub> = 2(1)\u00b2 + 3 x 1 = 2 + 3 = 5
\nS2<\/sub> = 2(2)\u00b2 + 3 x 2 = 8 + 6=14
\nS3<\/sub> = 2(3)\u00b2 + 3 x 3 = 18 + 9 = 27
\nNow, T2<\/sub> = S2<\/sub> – S1<\/sub> = 14 – 5 = 9
\nSecond term = 9<\/p>\n
\nSolution:<\/strong><\/span>
\nAP is a, 3a, 5a, …
\nHere, a = a, d = 2a
\n<\/p>\n
\nSolution:<\/strong><\/span>
\nAP 2, 7, 12, 17, …… 47
\nHere, a = 2, d = 7 – 2 = 5, l = 47
\nnth term from the end = l – (n – 1 )d
\n5th term from the end = 47 – (5 – 1) x 5 = 47 – 4 x 5 = 47 – 20 = 27<\/p>\n
\nSolution:<\/strong><\/span>
\nAP is 2, 7, 12, 17, …
\nHere, a = 2, d = 7 – 2 = 5
\nan<\/sub> = a + (n – 1) d = 2 + (n – 1) x 5 = 2 + 5n – 5 = 5n – 3
\nNow, a30<\/sub> = 2 + (30 – 1) x 5 = 2 + 29 x 5 = 2 + 145 = 147
\nand a20<\/sub> = 2 + (20 – 1) x 5 = 2 + 19 x 5 = 2 + 95 = 97
\na30<\/sub> – a20<\/sub> = 147 – 97 = 50<\/p>\n
\nSolution:<\/strong><\/span>
\nTn<\/sub> = 3n + 5
\nTn-1<\/sub> = 3 (n – 1) + 5 = 3n – 3 + 5 = 3n + 2
\nd = Tn<\/sub> – Tn-1<\/sub> = (3n + 5) – (3n + 2) = 3n + 5 – 3n – 2 = 3
\nCommon difference = 3<\/p>\n
\nSolution:<\/strong><\/span>
\nTn\u00a0<\/sub>= 7 – 4n
\nTn-1<\/sub> = 7 – 4(n – 1) = 7 – 4n + 4 = 11 – 4n
\nd = Tn<\/sub> – Tn-1<\/sub> = (7 – 4n) – (11 – 4n) = 7 – 4n – 11 + 4n = -4
\nd = -4<\/p>\n
\nSolution:<\/strong><\/span>
\nAP is \u221a8, \u221a18, \u221a32, …..
\n\u21d2 \u221a(4 x 2) , \u221a(9 x 2) , \u221a(16 x 2), ………
\n<\/p>\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nSolution:<\/strong><\/span>
\nAP is 21, 18, 15, …n
\nHere, a = 21, d = 18 – 21 = -3, l = 0
\nTn<\/sub> (l) = a + (n – 1) d
\n0 = 21 + (n – 1) x (-3)
\n0 = 21 – 3n + 3
\n\u21d2 24 – 3n = 0
\n\u21d2 3n = 24
\n\u21d2 n = 8 .
\n0 is the 8th term.<\/p>\n
\nSolution:<\/strong><\/span>
\nFirst n natural numbers are 1, 2, 3, 4, 5, …, n
\nHere, a = 1, d = 1
\n
\n<\/p>\n
\nSolution:<\/strong><\/span>
\nFirst n even natural numbers are 2, 4, 6, 8, 10, … n
\nHere, a = 2, d = 4 – 2 = 2
\n<\/p>\n
\nSolution:<\/strong><\/span>
\nIn an AP
\nFirst term (a) = p
\nand common difference (d) = q
\nT10<\/sub> = a + (n – 1) d = p + (10 – 1) x q = (p + 9q)<\/p>\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nSolution:<\/strong><\/span>
\n2p + 1, 13, 5p – 3 are in AP, then
\n13 – (2p + 1) = (5p – 3) – 13
\n\u21d2 13 – 2p – 1 = 5p – 3 – 13
\n\u21d2 12 – 2p = 5p – 16
\n\u21d2 5p + 2p = 12 + 16
\n\u21d2 7p = 28
\n\u21d2 p = 4
\nP = 4<\/p>\n
\nSolution:<\/strong><\/span>
\n(2p – 1), 7, 3p are in AP, then
\n\u21d2 7 – (2p – 1) = 3p – 7
\n\u21d2 7 – 2p + 1 = 3p – 7
\n\u21d2 7 + 1 + 7 = 3p + 2p
\n\u21d2 5p = 15
\n\u21d2 p = 3
\nP = 3<\/p>\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nSolution:<\/strong><\/span>
\n
\nd = T2<\/sub> – T1<\/sub> = 14 – 8 = 6
\nCommon difference = 6<\/p>\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nSolution:<\/strong><\/span>
\n<\/p>\n
\nSolution:<\/strong><\/span>
\n
\n<\/p>\n