RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions


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RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions





Question 1:





Solution:





(i) 9, 15, 21, 27, …





Here, 15 – 9 = 21 – 15 = 27 – 21 = 6 (which is constant)





Common difference is 6





Or d = 6





Next term = 27 + d = 27 + 6 = 33





(ii) 11, 6, 1, -4, …





Here, 6 – 11 = 1 – 6 = -5 -4 – 1 = -5 (which is constant)





d (common difference) = -5





Next term = -4 – 5 = -9





(iii) -1, -5 /6 , -2 /3 , -1 /2 , …………





-5/6 – (-1) = 1/6 and





-2/3 – (-5/6) = 1/6





d (common difference) = 1/6





Next term = -1/2 + 1/6 = -1/3





(iv) √2, √8, √18, √32, ……..





√8 – √2 = 2√2 – √2 = √2





√32 – √18 = 4√2 – 3√2 = √2





d (common difference) = √2





Next term = √32 + √2 = 4√2 + √2 = 5√2 = √50





(v) √20, √45, √80, √125, ……





√20, √45, √80, √125, ……





√45 – √20 = 3√5 – 2√5 = √5





√125 – √80 = 5√5 – 4√5 = √5





d (common difference) = √5





Next term = √125 + √5 = 5√5 + √5 = 6√5 or √180





Question 2:





Solution:





(i)Given: AP is 9, 13, 17, 21, ……





Here, first term = a = 9





Common difference = d = 13 – 9 = 4





an = a + (n-1)d





a20= 9 + (20-1)4





= 85





(ii) the 35th term of the AP 20, 17, 14, 11, ……





Given: AP is 20, 17, 14, 11, ……





Here, first term = a = 20





Common difference = d = 17 – 20 = -3





n = 35





an = a + (n-1)d





a35 = 20 + (35-1)(-3)





= -82





(iii) the 18th term of the AP √2, √18 , √50, √98, ………





Given: AP is √2, √18 , √50, √98, ………





or √2, 3√2 , 5√2, 7√2,….





Here, first term = a = √2





Common difference = d = 3√2 – √2 = 2√2





n = 18





an = a + (n-1)d





a18 = √2 + 34√2





= 35√2





(iv) the 9th term of the AP 3 /4 , 5 /4 , 7 /4 , 9 /4 , …….





Given: AP is 3 /4 , 5 /4 , 7 /4 , 9 /4 , …….





Here, first term = a = 3/4





Common difference = d = 5/4 – 3/4 = 1/2





n = 9





an = a + (n-1)d





a9 = 3/4 + (9 -1)1/2





= 19/4





(v) the 15th term of the AP -40, -15, 10, 35, ……





Given: AP is -40, -15, 10, 35, ……





Here, first term = a = -40





Common difference = d = -15 – (-40) = -15 + 40 = 25





n = 15





an = a + (n-1)d





a15 = -40 + (15 – 1)25





= 310





Question 3:





Solution:





rs aggarwal class 10 chapter 5 ex a 1




n = 25





an = a + (n-1)d





a25 = 5 + (25-1)(-1/2)





= -7





Question 4:





Solution:





If 2p – 1, 3p + 1, 11 are terms in AP, then





a2 – a1 = a3 – a…..(1)





From given:





a= 2p – 1





a2 = 3p + 1





a= 11





From (1), we get





(3p + 1) – (2p – 1) = 11 – (3p + 1)





3p + 1 – 2p + 1 = 11 – 3p – 1





p + 2 = 10 – 3p





4p = 8





p = 2





For p = 2, these terms are in AP.





Question 5:





Solution:





(i)AP is 5, 11, 17, 23, ……





Here, first term = a = 5





Common difference = d = 11 – 5 = 6





Now,





an = a (n – 1)d





= 5 + (n – 1) 6





= 5 + 6n – 6





= (6n – 1)





(ii) AP is 16, 9, 2, -5, ……





Here, first term = a = 16





Common difference = d = 9 – 16 = -7





an = a + (n – 1)d





= 16 + (n – 1) (-7)





= 16 – 7n + 7





= (23 – 7n)





Question 6:





Solution:





nth term of AP is 4n – 10 (Given)





Putting n = 1, 2, 3, 4, …, we get





At n = 1: 4n – 10 = 4 x 1 – 10 = 4 – 10 = -6





At n = 2: 4n – 10 = 4 x 2 – 10 = 8 – 10 = -2





At n = 3: 4n – 10 = 4 x 3 – 10 = 12 – 10 = 2





At n = 1: 4n – 10 = 4 x 4 – 10 = 16 – 10 = 6





We see that -6, -2, 2, 6,… are in AP





(i) first term = -6





(ii) Common difference = -2 – (-6) = 4





(iii) 16th term:





Using formula: an = a + (n – 1)d





Here n = 16





a16 = -6 + (16 – 1)4 = 54





Question 7:





Solution:





Given: AP is 6, 10, 14, 18,…, 174





Here, first term = a = 6





Common difference = d= 10 – 6 = 4





To find: the number of terms (n)





Last term = a + (n – 1)d





174 = 6 + (n – 1) 4





174 – 6 = (n – 1) 4





n – 1 = 168 /4 = 42





n = 42 + 1 = 43





There are 43 terms.





Question 8:





Solution:





Given: AP is 41, 38, 35,…, 8





Here, first term = a = 41





Last term = 8





Common difference = d = 38 – 41 = -3





To find: the number of terms (n)





Last term = a + (n – 1)d





8 = 41 + (n – 1)(-3)





8 – 41 = (n – 1)(-3)





n – 1 = 11





n = 11 + 1 = 12





There are 12 terms.





Question 9:





Solution:





Given: AP is 18, 31/2 , 13, …, -47





Here, first term = a = 18





Last term = -47





Common difference = d = -5/2





To find: the number of terms (n)





Last term = a + (n – 1)d





-47 = 18 + (n – 1)(-5/2)





-47 – 18 = (n – 1)(-5/2)





n = 27





There are 27 terms.





Question 10:





Solution:





Let nth term is 88.





AP is 3, 8, 13, 18, …





Here,





First term = a = 3





Common difference = d = 8 – 3 = 5





nth term of AP is an = a + (n – 1) d





Now,





88 = 3 + (n – 1)(5)





88 – 3 = (n – 1) x 5





n – 1 = 88 /5





or n = 17 + 1 = 18





Therefore: 88 is the 18th term.





Question 11:





Solution:





AP is 72, 68, 64, 60, …..





Let nth term is 0.





Here,





First term = a = 72





Common difference = d = 68 – 72 = -4





an = a + (n – 1)d





0 = 72 + (n – 1)(-4)





-72 = -4(n – 1)





n – 1 = 18





n = 18 + 1 = 19





Therefore: 0 is the 19th term.





Question 12:





Solution:





rs aggarwal class 10 chapter 5 ex a 2




Here,





First term = a = 5/6





Common difference = d = 1 – 5/6 = 1/6





Now: an = a + (n – 1)d





3 = 5/6 + (n – 1)1/6





Let nth term is 3





Now, an = a + (n-1)d





3 = 5/6 + (n-1)1/6





n -1 = 13





n = 13 + 1 = 14





Therefore, 3 is the 14th term.





Question 13:





Solution:





AP is 21, 18, 15, ……





Let nth term -81





Here, a = 21, d = 18 – 21 = -3





an = a + (n – 1)d





-81 = 21 + (n – 1)(-3)





-81 – 21 = (n – 1)(-3)





-102 = (n – 1)(-3)





n = 34 + 1 = 35





Therefore, -81 is the 35th term





Question 14:





Solution:





Given AP is 3, 8, 13, 18,…





First term = a = 3





Common difference = d = 8 – 3 = 5





And n = 20 and a20 be the 20th term, then





a20 = a + (n – 1)d





= 3 + (20 – 1) 5





= 3 + 95





= 98





The required term = 98 + 55 = 153





Now, 153 be the nth term, then





an = a + (n – 1)d





153 = 3 + (n – 1) x 5





153 – 3 = 5(n – 1)





150 = 5(n – 1)





n – 1 = 30





n = 31





Required term will be 31st term.





Question 15:





Solution:





AP is 5, 15, 25,…





First term = a = 5





Common difference = d = 15 – 5 = 10





Find 31st term:





a31 = a + (n – 1)d





= 5 + (31 – 1) 10





= 5 + 30 x 10





= 305





Required term = 305 + 130 = 435





Now, say 435 be the nth term, then





an = a + (n – 1)d





435 = 5 + (n – 1)10





435 – 5 = (n – 1)10





n – 1 = 43





n = 44





The required term will be 44th term.





Question 16:





Solution:





Let a be the first term and d be the common difference, then





rs aggarwal class 10 chapter 5 ex a 3




Question 17:





rs aggarwal class 10 chapter 5 ex a 4




a16 = 6 + (16 – 1)7 = 6 + 105 = 111





Therefore, midterm of the AP id 111.





Question 18:





Solution:





AP is 10, 7, 4, …..…, (-62)





a = 10,





d = 7 – 10 = -3, and





l = -62





Now, an = l = a + (n – 1)d





-62 = 10 + (n – 1) x (-3)





-62 – 10 = -3(n- 1)





-72 = -3(n – 1)





Or n = 24 + 1 = 25





Middle term = (25 + 1) /2 th = 13th term





Find the 13th term using formula, we get





a13 = 10 + (13 – 1)(-3) = 10 – 36 = -26





Question 19:





Solution:





Given AP is -4 /3 , -1, -2 /3 , …, 13 /3





rs aggarwal class 10 chapter 5 ex a 5




rs aggarwal class 10 chapter 5 ex a 6




Middle terms will be: (18/2)th + (18/2 + 1)th





= 9th + 10th term





Now,





a9 + a10 = a + 8d + a + 9d





= 2a + 17d





= 2(-4/3) + 17(1/3)





= 3





Question 20:





Solution:





Given: AP is 7, 10, 13,…, 184





a = 7, d = 10 – 7 = 3 and l = 184





nth term from the end = l – (n – 1)d





Now,





8th term from the end be





184 – (8 – 1)3 = 184 – 21 = 163





Question 21:





Solution:





Given: AP is 17, 14, 11, …,(-40)





a = 17, d = 14 – 17 = -3, l = -40





6th term from the end = l – (n – 1)d





= -40 – (6 – 1) (-3)





= -40 – (5 x (-3))





= -40 + 15





= -25





Question 22:





Solution:





Given AP is 3, 7, 11, 15, …





a = 3, d = 7 – 3 = 4





Let 184 be the nth term of the AP





an = a + (n – 1)d





184 = 3 + (n – 1) x 4





184 – 3 = (n – 1) x 4





181 /4 = n – 1





n = 181 /4 + 1 = 185/4 (Which is in fraction)





Therefore, 184 is not a term of the given AP.





Question 23:





Solution:





Given AP is AP 11, 8, 5, 2,…





Here a = 11, d = 8-11= -3





Let -150 be the nth term of the AP





an = a + (n – 1)d





-150 = 11 +(n-1)(-3)





or n = 164/3





Which is a fraction.





Therefore, -150 is not a term of the given AP.





Question 24:





Solution:





Let nth of the AP 121, 117, 113,… is negative. Let Tn be the nth term then





rs aggarwal class 10 chapter 5 ex a 7




Therefore, 32nd term will be the 1st negative term.





Question 25:





Solution:





rs aggarwal class 10 chapter 5 ex a 8




a = 20, d = -3/4





Let nth term be the 1st negative term of the AP





an < 0





an = a + (n – 1)d





rs aggarwal class 10 chapter 5 ex a 9




Therefore, 28th term will be the 1st negative term.





Solution:





Let us say a be the first term and d be the common difference of an AP





an = a + (n – 1)d





a7 = a + (7 – 1)d





= a + 6d = -4 …………(1)





And a13 = a + 12d = -16 …..…..(2)





Subtracting equation (1) from (2), we get





6d = -16 – (-4) = -12





From (1), a + 6d = -4





a + (-12) = -4





a = -4 + 12 = 8





a = 8, d = -2





AP will be 8, 6, 4, 2, 0, ……





Question 27:





Solution:





Let a be the first term and d be the common difference of an AP.





a4 = a + (n- 1)d





= a + (4 – 1)d





= a + 3d





Since 4th term of an AP is zero.





a + 3d = 0





or a = -3d ….(1)





Similarly,





a25 = a + 24d = -3d + 24d = 21d …(2)





a11 = a + 10d = -3d + 10d = 7d ….(3)





From (2) and (3), we have





a25 = 3 x a11





Hence proved.





Question 28:





Solution:





Sixth term of an AP is zero





that is a6 = 0





a + 5d = 0





a = -5 d





Now, a15 = a + (n – 1 )d





a + (15 – 1)d = -5d + 14d = 9d





and a33 = a + (n – 1 )d = a + (33 – 1)d = -5d + 32d = 27d





Now, a33 : a12





27d : 9d





3 : 1





Which shows that a33 = 3(a15)





Hence proved.





Question 29:





Solution:





Let a be the first term and d be the common difference of an AP.





an = a + (n – 1)d





a4 = a + (4 – 1)d = a + 3d





a + 3d = 11 ………(1)





Now, a5 = a + 4d and a7 = a + 6d





Now, a5 + a7 = a + 4d + a + 6d = 2a + 10d





2a + 10d = 34





a + 5d = 17 ……..(2)





Subtracting (1) from (2), we get





2d = 17 – 11 = 6





d = 3





The common difference = 3





Question 30:





Solution:





Let a be the first term and d be the common difference of an AP.





nth term = an = a + (n – 1)d





Given: 9th term of an AP is -32 and the sum of its 11th and 13th terms is -94





Now,





a9 = a + 8d = -32 …(1)





a11 = a + 10d





a13 = a + 12d





Sum of 11th and 13th terms:





a11 + a13 = a + 10d + a + 12d





-94 = 2a + 22d





or a + 11d =-47 …(2)





Subtracting (1) from (2) , we have





3d = -47 + 32 = -15





or d = -5





Common difference is -5.










Exercise 5B Page No: 264





Question 1:





Solution:





Given: (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.





So, (4k – 6) – (3k – 2) = (k + 2) – (4k – 6)





2(4k – 6) = (k + 2) + (3k – 2)





8k – 12 = 4k + 0





8k – 4k = 0 + 12





or k = 3





Question 2:





Solution:





Given: (5x + 2), (4x – 1) and (x + 2) are terms in AP.





So, d = (4x – 1) – (5x + 2) = (x + 2) – (4x – 1)





2(4x – 1) = (x + 2) + (5x + 2)





8x – 2 = 6x + 2 + 2





8x – 2 = 6x + 4





8x – 6x = 4 + 2





or x = 3





The value of x is 3.





Question 3:





Solution:





Given: (3y – 1), (3y + 5) and (5y + 1) are 3 consecutive terms of an AP.





So, (3y + 5) – (3y – 1) – (5y + 1) – (3y + 5)





2(3y + 5) = 5y + 1 + 3y – 1





6y + 10 = 8y





8y – 6y = 10





2y = 10





Or y = 5





The value of y is 5.





Question 4:





Solution:





Given: (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.





So, 2x – (x + 2) = (2x + 3) – 2x





2x – x – 2 = 2x + 3 – 2x





x – 2 = 3





x = 2 + 3 = 5





The value of x is 5.





Question 5:





Solution:





Assume that (a – b)², (a² + b²) and (a + b)² are in AP.





So, (a² + b²) – (a – b)² = (a + b)² – (a² + b²)





(a² + b²) – (a² + b² – 2ab) = a² + b² + 2ab – a² – b²





2ab = 2ab





Which is true.





Hence given terms are in AP.





Question 6:





Solution:





Let a – d, a, a + d are three numbers in AP.





So, their sum = a – d + a + a + d = 15





3a = 15





or a = 5





Again,





Their Product = (a – d) x a x (a + d) = 80





a(a² – d²) = 80





5(5² – d²) = 80





25 – d² = 16





d² = 25 – 16 = 9 = (±3)²





or d = ±3





d = 3 or d = -2





We have 2 conditions here:





At a = 5, d = 3





Numbers are: 2, 5 and 8





At a = 5 and d = -3





Numbers are : 8, 5, 2





Question 7:





Solution:





Let a – d, a, a + d are three numbers in AP.





their sum = a – d + a + a + d = 3





3a = 3





or a = 1





Again,





Their Product = (a – d) x a x (a + d) = -35





a(a² – d²) = -35





(1² – d²) = -35





or d = ±6





d = 6 or d = -6





We have 2 conditions here:





At a = 1, d = 6





Numbers are: -5, 1 and 7





At a = 1 and d = -6





Numbers are : 7, 1, -5





Question 8:





Solution:





Let a – d, a, a + d are three numbers in AP.





their sum = a – d + a + a + d = 24





3a = 24





or a = 8





Again,





Their Product = (a – d) x a x (a + d) = 440





a(a² – d²) = 440





8(8² – d²) = 440





or d = ±3





Numbers are : (5, 8, 11) or (11, 8, 5)





Question 9:





Solution:





Let a – d, a, a + d are three numbers in AP.





their sum = a – d + a + a + d = 21





3a = 21





or a = 7





Again,





Sum of squares = (a – d)^2 + a^2 +(a + d)^2 = 165





rs aggarwal class 10 chapter 5 ex b 1




Numbers are : (4, 7, 10) or (10, 7, 4)





Question 10:





Solution:





Sum of angles of a quadrilateral = 360°





Common difference = 10 = d (say)





If the first number be a, then the next four numbers will be





a, a + 10, a + 20, a + 30





As per definition:





a + a + 10 + a + 20 + a + 30 = 360°





4a + 60 = 360





4a = 300





or a = 75°





Other angles:





a + 10 = 75 + 10 = 85





a + 20 = 75 + 20 = 95





a + 30 = 75 + 30 = 105





Therefore, Angles are 75°, 85°, 95°, 105°





Question 11:





Solution:





Let a – 3d, a – d, a + d, a + 3d are the four numbers in AP.





Their sum = a – 3d + a – d + a + d + a + 3d = 28





Sum of their square = (a – 3d)^2 + (a – d)^2 + (a + d)^2 + (a + 3d)^2 = 216





rs aggarwal class 10 chapter 5 ex b 2




Numbers will be: (4, 6, 8,10) or ( 10, 8, 6, 4)





Question 12:





Solution:





Let a – 3d, a – d, a + d, a + 3d are the four numbers in AP.





Therefore:





rs aggarwal class 10 chapter 5 ex b 3




d = ±2





four parts are: a – 6, a – 2, a+2 amd a + 6





which implies,





Number are: (2, 6, 8, 10, 14) or (14, 10, 6, 2)





Question 13:





Solution:





Let a – d, a, a + d are the three terms in AP





So,





Sum = a – d + a + a + d = 48





3a = 48





Or a = 16





And,





a(a – d) = 4 (a + d) + 12





16 (16 – d) = 4(16 + d) + 12





256 – 16d = 64 + 4d + 12 = 4d + 76





256 – 76 = 4d + 16d





180 = 20d





d = 9





Which implies:





Numbers are : (7, 16, 25)










Exercise 5C Page No: 281





Question 1:





Solution:





Sum of n terms of AP formula:





rs aggarwal class 10 chapter 5 ex c 1……(1)





Where





First term = a





Common difference = d





Number of terms = n





(i) 2, 7, 12, 17,….. to 19 terms.





a = 2





d = 7 – 2 = 5





Using (1)





S19 = 19/2(2(2) + (19 – 1)5)





= (19)(4 + 90)/2





= (19 × 94)/2





= 893





Sum of 19 terms of this AP is 893.





(ii) 9, 7, 5, 3,…..to 14 terms.





a = 9





d = 7 – 9 = -2





Using (1)





S14 = 14/2 [2(9) + (14 – 1)(-2)]





= (7)(18 – 26)





= – 56





Sum of 14 terms of this AP is – 56.





(iii) -37, -33, -29,….12 terms.





a = -37





d = (-33) – (-37) = 4





Using (1)





S12 = 12/2 [2(-37) + (12 – 1)(4)]





= (6)(-74 + 44)





= 6 × (-30)





= – 180





Sum of 12 terms of this AP is – 180.





(iv) 1/15 ,1/12,1/10,…to 11 terms





a = 1/15





d = (1/12) – (1/15) = 1/60





Using (1)





S11 = 11/2 [2(1/15) + (11 – 1)(1/60)]





= (11/2) × [(2/15) + (1/6)]





=33/20





Sum of 11 terms of this AP is 33/20.





(v) 0.6, 1.7, 2.8,…. to 100 terms.





a = 0.6





d = 1.7 – 0.6 = 1.1





Using (1)





S100 = 100/2 [2(0.6) + (100 – 1)(1.1)]





= (50) × [1.2 + (99 × 1.1) ]





= 50 × 110.1





= 5505





Sum of 100 terms of this AP is 5505.





Question 2:





Solution:





(i) 7 + 10 1/2 + 14 + … + 84





First term = a = 7





Common difference = d = (21/2) – 7 = (7/2)





Last term = l = 84





Now, using formula:





84 = a + (n – 1)d





84 = 7 + (n – 1)(7/2)





77 = (n – 1)(7/2)





154 = 7n – 7





7n = 161





n = 23





Thus, there are 23 terms in AP.





Now,





Find Sum of these 23 terms:





S23 = 23/2 [2(7) + (23 – 1)(7/2)]





= (23/2) [14 + (22)(7/2) ]





= (23/2) [91]





= 2093/2





Sum of 23 terms of this AP is 2093/2.





(ii) 34 + 32 + 30 + … + 10





First term = a = 34





Common difference = d = 34 – 32 = – 2





Last term = l = 10





Now, using formula:





10 = a + (n – 1)d





10 = 34 + (n – 1)(-2)





10 – 34 = (n – 1)(-2)





n = 13





Thus, there are 13 terms in AP.





Now,





Find Sum of these 13 terms:





S13 = 13/2 [2(34) + (13 – 1)(-2)]





= (13/2) [68 + (12)(-2)]





= (13/2) x 44





= 286





Sum of 13 terms of this AP is 286.





(iii) (-5) + (-8) + (-11) + … + (-230)





First term = a = -5





Common difference = d = – 8 – (-5) = – 3





Last term = l = -230





Now, using formula:





-230 = a + (n – 1)d





-230 = -5 + (n – 1)(-3)





– 230 + 5 = (n – 1)(-3)





n = 76





Thus, there are 76 terms in AP.





Now,





Find Sum of these 76 terms:





S76 = 76/2 [2(-5) + (76 – 1)(-3)]





= 38 × [(-10) + (75)(-3)]





= – 8930





Sum of 76 terms of this AP is -8930.





(iv) 5 + (-41) + 9 + (-39) + 13 + (-37) + 17 + … + (-5) + 81 + (-3)





The given series is combination of two AP’s.





Let A1 = 5 + 9 +13 + ……+ 77 + 81





and A2 = -41 – 39 – 37 – ……(-3)





For A1:





First term = a = 5





Common difference = d = 2





Last term = l = 81





Now, using formula:





81 = 5 + (n – 1)4





n = 20





Thus, there are 20 terms in AP.





Now,





Find Sum of these 76 terms:





S20 = n/2 [a + l]





= 20/2 (5 + 81)





= 860





Sum of 20 terms of this AP is 860.





For A2:





First term = a = -41





Common difference = d = 2





Last term = l = -3





Now, using formula:





-3 = -41 + (n – 1)2





n = 20





Thus, there are 20 terms in AP.





Now,





Find Sum of these 76 terms:





S20 = n/2 [a + l]





= 20/2 (-41- 3)





=-440





Sum of 20 terms of this AP is -440.





Therefore, Sum of total terms: 860 – 440 = 420.





Question 3: 





Solution:





Given: an = 5 – 6n





Find some of the terms of AP:





Put n = 1, we get a1 = – 1 = first term





Put n = 2, we get a2 = – 7 = second term





Common difference = d = a2 – a1 = – 7 – (-1) = – 6





Sum of first n terms:





Sn= n/2 [2a + (n – 1)d]





= n/2[- 2 + (n – 1)(-6)]





= n(2 – 3n)





sum of first 20 terms:





S20 = 20/2[2(-1) + (20 – 1)(-6)]





= 10 [ – 2 – 114]





= – 1160





Sum of its first 20 terms of AP is -1160.





Question 4:





Solution:





Given: Sn = 3n² + 6n





rs aggarwal class 10 chapter 5 ex c 2




Question 5:





Solution:





Sn = 3n² – n





S1 = 3(1)² – 1 = 3 – 1 = 2





S2 = 3(2)² – 2 = 12 – 2 = 10





a1 = 2





a2 = 10 – 2 = 8





(i) an = a + (n – 1) d





= 2 + (n – 1) x 6





= 2 + 6n – 6





= 6n – 4





(ii) First term = 2





(iii) Common difference = 8 – 2 = 6





Question 6: 





Solution:





rs aggarwal class 10 chapter 5 ex c 3




rs aggarwal class 10 chapter 5 ex c 4




(ii)





rs aggarwal class 10 chapter 5 ex c 5




Question 7: 





Solution:





Let a be first term and d be the common difference of an AP.





mth term = 1/n





so,





am = a + (m – 1) d = 1/n ……….(1)





nth term = 1/m





an = a + (n – 1) d = 1/m ……….(2)





Subtract (2) from (1)





rs aggarwal class 10 chapter 5 ex c 6




Sum of first mn terms:





Smn = mn/2 [2(1/mn) + (mn-1) (1/mn)]





= mn/2 [1/mn + 1]





= (1+mn)/2





Question 8:





Solution:





AP is 21, 18, 15,…





a = 21,





d = 18 – 21 = -3





Sum of terms = Sn = 0





(n/2) [2a + (n – 1)d] = 0





(n/2) [2(21) + (n – 1)(-3)] = 0





(n/2) [45 – 3n] = 0[45 – 3n] = 0





n = 15 (number of terms)





Thus, 15 terms of the given AP sums to zero.





Question 9: 





Solution:





AP is 9, 17, 25,…





a = 9, d = 17 – 9 = 8





Sum of terms = Sn = 636





(n/2) [2a + (n – 1)d] = 636





(n/2) [2(9) + (n – 1)(8)] = 636





(n/2)[10 + 8n] = 636





4n2 + 5n – 636 = 0 (which is a quadratic equation)





(n- 12)(4n + 53) = 0





Either (n- 12) = 0 or (4n + 53) = 0





n = 12 or n = – 53/4





Since n can’t be negative and fraction, so





n = 12





Number of terms = 12 terms.





Question 10: 





Solution:





AP is 63, 60, 57, 54,…





Here, a = 63, d = 60 – 63 = -3 and sum = Sn = 693





rs aggarwal class 10 chapter 5 ex c 7




Which is a quadratic equation





rs aggarwal class 10 chapter 5 ex c 8




Which shows that, 22th term of AP is zero.





Number of terms are 21 or 22. So there will be no effect on the sum.





Question 11:





Solution:









Here, a = 20, d = -2/3 and sum = sn = 300 (for n number of terms)





rs aggarwal class 10 chapter 5 ex c 9




rs aggarwal class 10 chapter 5 ex c 10




n(n-25) – 36(n-25) = 0





(n – 25)(n – 36) = 0





Either (n – 25) = 0 or (n – 36) = 0





n = 25 or n = 36





For n = 25





rs aggarwal class 10 chapter 5 ex c 11




= 300





Which is true.





Result is true for both the values of n





So both the numbers are correct.





Therefore, Sum of 11 terms is zero. (36-25 = 11)





Question 12:





Solution:





Odd numbers between 0 and 50 are 1, 3, 5, 7, 9, …, 49





Here, a = 1, d = 3 – 1 = 2, l = 49





rs aggarwal class 10 chapter 5 ex c 12




Sum of odd numbers:





= n/2(a + l)





= 25/2 (l + 49)





= 25/2(50)





= 625





Question 13: 





Solution:





Natural numbers between 200 and 400 which are divisible by 7 are 203, 210, 217,……,399.





Here,AP is 203, 210, 217,……,399





First term = a = 203,





Common difference = d = 7 and





Last term = l = 399





We know that, l = a + (n-1)d





rs aggarwal class 10 chapter 5 ex c 13




Sum of all 29 terms is 8729.





Question 14: 





Solution:





First forty positive integers which are divisible by 6 are





6, 12, 18, 24,…… to 40 terms





Here, a = 6, d = 12 – 6 = 6, and n = 40.





Sum of 40 terms:





rs aggarwal class 10 chapter 5 ex c 14




Question 15:





Solution:





First 15 multiples of 8 are as given below:





8, 16, 24, 32,….. to 15 terms





Here, a = 8, d = 16 – 8 = 8, and n = 15





Sum of 15 terms:





rs aggarwal class 10 chapter 5 ex c 15




Question 16: 





Solution:





Multiples of 9 lying between 300 and 700 = 306, 315, 324, 333, …, 693





Here, a = 306, d = 9, and l = 693





We know that, l = a + (n-1)d





rs aggarwal class 10 chapter 5 ex c 16




There are 44 terms.





Find the sum of 44 terms:





rs aggarwal class 10 chapter 5 ex c 17




Question 17:





Solution:





3-digit natural numbers: 100, 101, 102,……., 999.





3-digit natural numbers divisible by 13:





104, 117, 130,………, 988





Which is an AP.





Here, a = 104, d = 13, l = 988





l = a + (n – 1) d





988 = 104 + (n-1)13





n = 69





There are 69 terms.





Find the sum of 69 terms:





Sn = n/2(a + l)





= 69/2(104 + 988)





= 37674





Question 18: 





Solution:





Even natural numbers: 2, 4, 6, 8, 10, …





Even natural numbers divisible by 5: 10, 20, 30, 40, … to 100 terms





Here, a = 10, d = 20 – 10 = 10, and n = 100





Sn = n/2(2a + (n-1)d)





= 100/2 [2 x 10 + (100-1)10]





= 50500





Question 19:





(4 – 1/n) + (4-2/n) + (4– 3/n) + ….





Solution:





Given sum can be written as (4 + 4 + 4 + 4+….) – (1/n, 2/n, 3/n, …..)





Now, We have two series:





First series: = 4 + 4 + 4 + …… up to n terms





= 4n





Second series: 1/n, 2/n, 3/n, …..





Here, first term = a = 1/n





Common difference = d = (2/n) – (1/n) = (1/n)





Sum of n terms formula:





Sn = n/2[2a + (n – 1)d]





Sum of n terms of second series:





Sn = n/2 [2(1/n) + (n – 1)(1/n)]





= n/2 [(2/n) + 1 – (1/n)]





= (n + 1)/2





Hence,





Sum of n terms of the given series = Sum of n terms of first series – Sum of n terms of second series





= 4n – (n + 1)/2





= (8n – n – 1)/2





= 1/2 (7n – 1)





Question 20: 





Solution:





Let a be the first term and d be the common difference of the AP.





rs aggarwal class 10 chapter 5 ex c 18




rs aggarwal class 10 chapter 5 ex c 19




Which implies: a = 1 and d = 5





Therefore, the AP is 1, 6, 11, 16,……





Question 21: 





Solution:





Let d be the common difference.





Given:





first term = a = 2





last term = l = 29





Sum of all the terms = Sn = 155





Sn = n/2[a + l]





155 = n/2[2 + 29]





n = 10





There are 10 terms in total.





Therefore, 29 is the 10th term of the AP.





Now, 29 = a + (10 – 1)d





29 = 2 + 9d





27 = 9d





d = 3





The common difference is 3.





Question 22:





Solution:





Let d be the common difference.





Given:





first term = a = -4





last term = l = 29





Sum of all the terms = Sn = 150





Sn = n/2[a + l]





150 = n/2[-4 + 29]





n = 12





There are 12 terms in total.





Therefore, 29 is the 12th term of the AP.





Now, 29 = -4 + (12 – 1)d





29 = -4 + 11d





d = 3





The common difference is 3.





Question 23:





Solution:





Let n be the total number of terms.





Given:





First term = a = 17





Last term = l = 350





Common difference = d = 9





l = a + (n-1)d





350 = 17 + (n-1)9





n = 38





Again,





Sn = n/2[a + l]





= 38/2[17 + 350]





= 6973





There are 38 terms in total and their sum is 6973.





Question 24:





Solution:





Let n be the total number of terms and d be the common difference.





Given:





first term = a = 5





last term = l = 45





Sum of all terms = Sn = 400





Sn = n/2[a + l]





400 = n/2[5 + 45]





n/2 = 400/50





n = 16





There are 16 terms in the AP.





Therefore, 45 is the 16th term of the AP.





45 = a + (16 – 1)d





45 = 5 + 15d





40 = 15d





15d = 40





d = 8/3





Common difference = d = 8/3





Common difference is 8/3 and the number of terms are 16.





Question 25:





Solution:





Let n be the total number of terms and d be the common difference.





Given:





first term = a = 22





nth term = -11





Sum of all terms = Sn = 66





Sn = n/2[a + l]





66 = n/2[22 + (-11)]





n = 12





There are 12 terms in the AP.





Since nth term is -11, so





an = a + (n – 1)d





-11 = 22 + (12-1)d





d = -3





Therefore, Common difference is -3 and the number of terms are 12.










Exercise 5D Page No: 289





Question 1: 





Solution:





Given: (3y – 1), (3y + 5) and (5y+ 1) are in AP





So, (3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)





2 (3y + 5) = (5y + 1) + (3y – 1)





6y + 10 = 8y





8y – 6y = 10





2y = 10





Or y = 5





The value of y is 5.





Question 2:





Solution:





Given: k, (2k – 1) and (2k + 1) are the three successive terms of an AP.





So, (2k – 1) – k = (2k + 1) – (2k – 1)





2 (2k – 1) = 2k + 1 + k





4k – 2 = 3k + 1





4k – 3k = 1 + 2





or k = 3





The value of k is 3.





Question 3:





Solution:





Given: 18, a, (b – 3) are in AP





a – 18 = b – 3 – a





a + a – b = -3 + 18





2a – b = 15





Question 4:





Solution:





Given: a, 9, b, 25 are in AP.





So, 9 – a = b – 9 = 25 – b





b – 9 = 25 – b





b + b = 22 + 9 = 34





or b = 17





And,





a – b = a – 9





9 + 9 = a + b





a + b = 18





a + 17 = 18





or a = 1





Answer: a = 18, b= 17





Question 5:





Solution:





Given: (2n – 1), (3n + 2) and (6n – 1) are in AP





So, (3n + 2) – (2n – 1) = (6n – 1) – (3n + 2)





(3n + 2) + (3n + 2) = 6n – 1 + 2n – 1





6n + 4 = 8n – 2





8n – 6n = 4 + 2





Or n = 3





Numbers are:





2 x 3 – 1 = 5





3 x 3 + 2 = 11





6 x 3 – 1 = 17





Answer: (5, 11, 17) are required numbers.





Question 6:





Solution:





3-digit natural numbers: 100, 101,…….. 990 and





3-digit natural numbers divisible by 7: 105, 112, 119, 126, …, 994





Here, a = 105, d= 7, l = 994





an = (l) = a + (n – 1) d





994 = 105 + (n – 1) x 7





994 – 105 = (n – 1) 7





n – 1 = 127





n = 128





Answer: There are 128 required numbers.





Question 7:





Solution:





3-digit numbers: 100, 101,…….,999





3-digit numbers divisible by 9 : 108, 117, 126, 135, …, 999





Here, a = 108, d= 9, l = 999





an (l) = a + (n – 1) d





999 = 108 + (n – 1) x 9





(n – 1) x 9 = 999 – 108 = 891





n – 1 = 99





n = 100





Question 8:





Solution:





Sum of first m terms of an AP = 2m² + 3m (given)





Sm = 2m² + 3m





Sum of one term = S1 = 2(1)² + 3 x 1 = 2 + 3 = 5 = first term





Sum of first two terms = S2 = 2(2)² + 3 x 2 = 8 + 6=14





Sum of first three terms = S3 = 2(3)² + 3 x 3 = 18 + 9 = 27





Now,





Second term = a2 = S2 – S1 = 14 – 5 = 9





Question 9: 





Solution:





AP is a, 3a, 5a,…..





Here, a = a, d = 2a





Sum = Sn = n/2 [2a + (n-1)d]





= n/2[2a + 2an – 2a]





=an2





Question 10:





Solution:





Given AP is 2, 7, 12, 17, …… 47





Here, a = 2, d = 7 – 2 = 5, l = 47





nth term from the end = l – (n – 1 )d





5th term from the end = 47 – (5 – 1) 5 = 47 – 4 x 5 = 27





Question 11:





Solution:





Given AP is 2, 7, 12, 17,……..





Here, a = 2, d = 7 – 2 = 5





Now,





an = a + (n – 1) d = 2 + (n – 1) 5 = 5n – 3





a30 = 2 + (30 – 1) 5 = 2 + 145 = 147 and





a20 = 2 + (20 – 1) 5 = 2 + 95 = 97





a30 – a20 = 147 – 97 = 50





Question 12:





Solution:





Nth term = an = 3n + 5 (given)





a(n-1) = 3 (n – 1) + 5 = 3n + 2





Common difference = d = an – a(n-1)





= (3n + 5) – (3n + 2)





= 3n + 5 – 3n – 2





= 3





Therefore, common difference is 3.





Question 13:





Solution:





Nth term = an = 7 – 4n





a(n-1) = 7 – 4(n – 1) = 11 – 4n





Common difference = d = an – a(n-1)





= (7 – 4n) – (11 – 4n)





= 7 – 4n – 11 + 4n





= -4





Therefore, common difference is -4.





Question 14:





Solution:





Given AP is √8, √18, √32,……..





Above AP can be written as:





2√2 , 3√2, 4 √2, ……





Here a = 2√2 and d = √2





Next term = 4√2 + √2 = 5√2 = √50





Question 15:





Solution:





Given AP is √2, √8, √18,….





Can be written as:





√2, 2√2, 3√2,…..





First term = √2





Common difference = 2√2 – √2 = √2





Next term = 3√2 + √2 = 4√2 = √32





Question 16:
Solution:





Given AP is 21, 18, 15,….





First term = a = 21





Common difference = d = 18-21 = -3





Last term = l = 0





l = a + (n – 1) d





0 = 21 + (n – 1)(-3)





0 = 21 – 3n + 3





24 – 3n = 0





Or n = 8





Answer: Zero is the 8th term.





Question 17:





Solution:





First n natural numbers: 1, 2, 3, 4, 5, …, n





Here, a = 1, d = 1





Sum = Sn = n/2 [2a + (n-1)d]





= n/2 [2(1) + (n-1)(1)]





= n(n+1)/2





Question 18:





Solution:





First n even natural numbers: 2, 4, 6, 8, 10,….,n





Here, a = 2, d = 4 – 2 = 2





Sum = Sn = n/2 [2a + (n-1)d]





= n/2 [2(2) + (n-1)(2)]





= n(n+1)





Question 19: 





Solution:





Given:





First term =a = p and





Common difference = d =q





Now,





a10 = a + (n – 1) d





= p + (10 – 1)q





= (p + 9q)





Question 20:





Solution:





AP terms: 4/5, a, 2 (given)





Then,





a – 4/5 = 2 – a





a = 7/5





Question 21:





Solution:





Given, 2p + 1, 13, 5p – 3 are in AP





Then,





13 – (2p + 1) = (5p – 3) – 13





13 – 2p – 1 = 5p – 3 – 13





12 – 2p = 5p – 16





p = 4





The value of p is 4.





Question 22:





Solution:





Given, (2p – 1), 7, 3p are in AP





Then,





7 – (2p – 1) = 3p – 7





7 – 2p + 1 = 3p – 7





5p = 15





p = 3





The value of p is 3.





Question 23:





Solution:





Sum of first p terms = Sp = (ap² + bp)





Sum of one term = S1 = a(1)² + b(1) = a+b = first term





Sum of first two terms = S2 = a(2)² + b x 2 = 4a + 2b





We know that, second term = a2 = S2 – S1





= (4a + 2b) – (a + b)





= 3a + b





Now, d = a2 – a1





= 3a + b – (a+b)





= 2a





Answer: Common difference is 2a.





Question 24:





Solution:





Sum of first n terms = Sn = (3n² + 5n)





Sum of one term = S1 = 3(1)² + 5(1) = 8 = first term





Sum of first two terms = S2 = 3(2)² + 5(2) = 22





We know that, second term = a2 = S2 – S1





=22 – 8





= 14





=> a2 = 14





Now, d = a2 – a1





= 14 – 8 = 6





Answer: Common difference is 6.





Question 25:





Solution:





Let a be the first term and d be the common difference.





Given:





4th term = a4 = 9





Sum of 6th and 13th terms = a6 + a13 = 40





Now,





a4 = a + (4-1)d





9 = a + 3d





a = 9 – 3d ….(1)





And





a6 + a13 = 40





a + 5d + a + 12d = 40





2a +17d = 40





2(9 – 3d) + 17d = 40 (using (1))





d = 2





From (1): a = 9 – 6 = 3





Required AP = 3,5,7,9,…..





Question 26: 





Solution:





Given: a27 – a7 = 84[a + 26d] – [a + 6d] = 84





20d = 84





d = 4.2





Question 27:





Solution:





Given: 1 + 4 + 7 + 10 + … + x = 287





Here a = 1, d = 3 and Sn = 287





Sum = Sn = n/2 [2a + (n-1)d]





287 = n/2 [2 + (n-1)3]





574 = 3n2 – n





Which is a quadratic equation.





Solve 3n2 – n – 574 = 0





3n2 – 42n + 41n – 574 = 0





3n(n – 14) + 41(n-14) = 0





(3n + 41)(n-14) = 0





Either (3n + 41) = 0 or (n-14) = 0





n = -41/3 or n = 14





Since number of terms cannot be negative, so result is n = 14.





=> Total number of terms in AP are 14.





Which shows, x = a14





or x = a + 13d





or x = 1 + 39





or x = 40





The value of x is 40.





Complete RS Aggarwal Solutions Class 10





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RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Reviewed by Admin on November 15, 2020 Rating: 5

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