Arithmetic Progression

 Problem : 1 [ Arithmetic Progression ]       Solve this type of problem 1. For given arithemetic progression series 7,3,-1,-5,-9 ,... find 10 th term and addition of first 10 th terms. Show Answer A: Here first term a = 7,d = 3 - 7 = -4We know that, f(n) = a + (n - 1)df(10) = 7 + (10 - 1)(-4)= 7 + (-36)= -29We know that, S_n = n/2 [2a + (n - 1)d]:. S_10 = 10/2 * [2(7) + (10 - 1)(-4)]= 5 * [14 + (-36)]= 5 * [-22]= -110Hence, 10^(th) term of the given series is -29 and sum of first 10^(th) term is -110

 Problem : 2 [ Arithmetic Progression ]       Solve this type of problem 2. For arithemetic progression f( 5 ) = 56 , f( 8 ) = 86 then find f( 10 ) and S( 10 ). Show Answer A: We know that, f(n) = a + (n - 1)df(5) = 56=> a + (5 - 1)d = 56=> a + 4d = 56 ->(1)f(8) = 86=> a + (8 - 1)d = 86=> a + 7d = 86 ->(2)Solving (1) and (2), we get a = 16 and d = 10We know that, f(n) = a + (n - 1)df(10) = 16 + (10 - 1)(10)= 16 + (90)= 106We know that, S_n = n/2 [2a + (n - 1)d]:. S_10 = 10/2 * [2(16) + (10 - 1)(10)]= 5 * [32 + (90)]= 5 * [122]= 610

 Problem : 3 [ Arithmetic Progression ]       Solve this type of problem 3. For arithemetic progression f( 5 ) = 25 , f( 11 ) = 49 , then find n such that f(n) = 105 . Show Answer A: We know that, f(n) = a + (n - 1)df(5) = 25=> a + (5 - 1)d = 25=> a + 4d = 25 ->(1)f(11) = 49=> a + (11 - 1)d = 49=> a + 10d = 49 ->(2)Solving (1) and (2), we get a = 9 and d = 4Let n be the term such that f(n) = 105We know that, f(n) = a + (n - 1)d9 + (n - 1)(4) = 105(n - 1)(4) = 96n - 1 = 24n = 25

 Problem : 4 [ Arithmetic Progression ]       Solve this type of problem 4. For arithemetic progression S( 33 ) = 198 , then find f( 17 ). Show Answer A: We have given S_33 = 198 and we have to find f(17) = ?We know that, S_n = n/2 [2a + (n - 1)d]S_33 = 33/2 [2a + (33 - 1)d] = 198=> 33 [a + 16d] = 198=> a + 16d = 6 ->(1)We know that, f(n) = a + (n - 1)df(17) = a + (17 - 1)d=> f(17) = a + 16d=> f(17) = 6 (Using (1))

 Problem : 5 [ Arithmetic Progression ]       Solve this type of problem 5. For arithemetic progression f( 17 ) = 6 , then find S( 33 ). Show Answer A: We have given f(17) = 6 and we have to find S_33 = ?We know that, f(n) = a + (n - 1)df(17) = a + (17 - 1)d=> a + 16d = 6 ->(1)We know that, S_n = n/2 [2a + (n - 1)d]S_33 = 33/2 * [2a + (33 - 1)d]=> S_33 = 33 * [a + 16 d]=> S_33 = 33 * 6 (Using (1))=> S_33 = 198

 Problem : 6 [ Arithmetic Progression ]       Solve this type of problem 6. For arithemetic progression f( 7 ) = 13 , S( 14 ) = 203 , then find f( 10 ) and S( 8 ). Show Answer A: We have given f(7) = 13, S_14 = 203 and we have to find f(10) = ? and S(8) = ?We know that, f(n) = a + (n - 1)df(7) = 13=> a + (7 - 1)d = 13=> a + 6d = 13 ->(1)We know that, S_n = n/2 [2a + (n - 1)d]S_14 = 203=> 14/2 * [2a + (14 - 1)d] = 203=> 7 * [2a + 13d] = 203=> 2a + 13d = 29 ->(2)Solving (1) and (2), we get a = -5 and d = 3We know that, f(n) = a + (n - 1)df(10) = -5 + (10 - 1)(3)= -5 + (27)= 22We know that, S_n = n/2 [2a + (n - 1)d]:. S_8 = 8/2 * [2(-5) + (8 - 1)(3)]= 4 * [-10 + (21)]= 4 * [11]= 44

 Problem : 7 [ Arithmetic Progression ]       Solve this type of problem 7. For arithemetic progression addition of 3 terms is 27 and their multiplication is 648 , then that nos. Show Answer A: Let the terms are a-d, a, a+dAddition of this terms is 27=> (a-d) + a + (a+d) = 27=> 3 a = 27=> a = 27/3 = 9Multiplication of this terms is 648=> (a-d) × a × (a+d) = 648=> (9-d) × 9 × (9+d) = 648=> (81 - d^2) = 72=> d^2 = 9=> d = +- 3d = +3 => Required terms : 9 - 3, 9, 9 + 3 => 6, 9, 12d = -3 => Required terms : 9-(-3), 9, 9-3 => 12, 9, 6

 Problem : 8 [ Arithmetic Progression ]       Solve this type of problem 8. For arithemetic progression addition of first 17 terms is 24 and addition of first 24 terms is 17 , then find addition of first 41 terms. Show Answer A: We know that, S_n = n/2 [2a + (n - 1)d]S_17 = 17/2 * [2a + (17 - 1)d] = 24=> 17/2 * [2a + 16d] = 24=> 2a + 16 d = 2.8235 ->(1)We know that, S_n = n/2 [2a + (n - 1)d]S_24 = 24/2 * [2a + (24 - 1)d] = 17=> 24/2 * [a + 23d] = 17=> 2a + 23d = 1.4167 ->(2)Solving 7 d = -1.4069=> d = -0.201From (1) => 2a + 16d = 2.8235=> 2a = 2.8235 - 16d=> 2a = 2.8235 - 16 × -0.201=> 2a = 2.8235 - -3.2157=> 2a = 6.0392=> a = 3.0196We know that, S_n = n/2 [2a + (n - 1)d]:. S_41 = 41/2 * [2(3.0196) + (41 - 1)(-0.201)]= 41/2 * [6.0392 + (-8.0392)]= 41/2 × -2= -41

 Problem : 9 [ Arithmetic Progression ]       Solve this type of problem 9. For arithmetic progression Sm = n and Sn = m then prove that Sm+n = -(m - n) Show Answer A: For arithmetic progression, S_n = n/2 [ 2a + (n - 1) d ]Now, S_m = n:. m/2 [ 2a + (m - 1) d ] = n:. [ 2a + (m - 1) d ] = (2n)/m ->(1)Now, S_n = m:. n/2 [ 2a + (n - 1) d ] = m:. [ 2a + (n - 1) d ] = (2m)/n ->(2)(1) - (2) =>(m - 1) d - (n - 1) d = (2n)/m - (2m)/n:. (m - n) d = (2 (n^2- m^2))/(mn) :. (m - n) d = (-2 (m - n)(m + n))/(mn) :. d = (-2 (m + n))/(mn) ->(3)Now, S_(m+n) = (m + n)/2 [ 2a + (m + n - 1) d ]= (m + n)/2 [ 2a + (m - 1) d + nd ]= (m + n)/2 [ (2n)/m + n ( (-2(m+n))/(mn) ) ] (because from (1) and (3))= (m + n)/2 [ (2n)/m + (-2(m+n))/m ] = (m + n)/2 [ (2n - 2m - 2n)/m ]= (m + n)/2 [ (- 2m)/m ]= (m + n)/2 [ -2 ]= -(m + n) (Proved)

 Problem : 10 [ Arithmetic Progression ]       Solve this type of problem 10. For arithmetic progression Sm = n and Sn = m then prove that Sm-n = (m - n)(1 + 2n / m) Show Answer A: For arithmetic progression, S_n = n/2 [ 2a + (n - 1) d ]Now, S_m = n:. m/2 [ 2a + (m - 1) d ] = n:. [ 2a + (m - 1) d ] = (2n)/m ->(1)Now, S_n = m:. n/2 [ 2a + (n - 1) d ] = m:. [ 2a + (n - 1) d ] = (2m)/n ->(2)(1) - (2) =>(m - 1) d - (n - 1) d = (2n)/m - (2m)/n :. (m - n) d = (2 (n^2 - m^2))/(mn):. d = (-2 (m + n))/(mn) -->(3)Now, S_(m-n) = (m - n)/2 [ 2a + (m - n - 1) d ]= (m - n)/2 [ 2a + (m - 1) d - nd ]= (m - n)/2 [ (2n)/m - n ( (-2(m+n))/(mn) ) ] (because from (1) and (3))= (m - n)/2 [ (2n)/m - (-2(m+n))/m ] = (m - n)/2 [ ((2n + 2m + 2n))/m ]= (m - n)/2 [ ((2m + 4n))/m ]= (m - n)/2 [ (2(m + 2n))/m ]= (m - n) [ (m + 2n)/m ]= (m - n)(1 + (2n)/m) (Proved)

 Problem : 11 [ Arithmetic Progression ]       Solve this type of problem 11. The ratio of two arithemetic progression series is 3x+5 : 4x-2 , then find the ratio of their 10 th term. Show Answer A: Let two series be a, a+d, a+2d, ... and A, A+D, A+2D, ...Now, S_n / S_(n') = (3x+5) / (4x-2)=> (n/2 [2a + (n - 1)d]) / (n/2 [2A + (n - 1)D]) = (3x+5) / (4x-2)=> [2a + (n - 1)d] / [2A + (n - 1)D] = (3x+5) / (4x-2) ->(1)We know that, f(n) = a + (n - 1)df(n)/(F(n)) = [a + (n - 1)d] / [A + (n - 1)D]=> f(10)/(F(10)) = [a + (10 - 1)d] / [A + (10 - 1)D]=> f(10)/(F(10)) = [a + 9d] / [A + 9D]=> f(10)/(F(10)) = [2a + 18d] / [2A + 18D] ->(2)Using (1) and (2), it is clear that if we put n = 19 in (1) we get the ratio of (2)=> f(10)/(F(10)) = [2a + 18d] / [2A + 18D]=> f(10)/(F(10)) = (3x+5) / (4x-2), where n=19=> f(10)/(F(10)) = 62 / 74=> f(10)/(F(10)) = 31/37

 Problem : 12 [ Arithmetic Progression ]       Solve this type of problem 12. Find the sum of all natural nos between 100 to 200 and which are divisible by 4 . Show Answer A: Numbers between 100 and 200 divisible by 4 are 100, 104, 108 ...Which are in arithmetic progression.In which a=100 and d=4Let n be the term such that f(n) = 200We know that, f(n) = a + (n - 1)d100 + (n - 1)(4) = 200(n - 1)(4) = 100n - 1 = 25n = 26We know that, S_n = n/2 [2a + (n - 1)d]:. S_26 = 26/2 * [2(100) + (26 - 1)(4)]= 13 * [200 + (100)]= 13 * [300]= 3900

 Problem : 13 [ Arithmetic Progression ]       Solve this type of problem 13. Find the sum of all natural nos between 100 to 200 and which are not divisible by 4 . Show Answer A: Required Addition = (100 + 101 + 102... + 200) - (100 + 104 + 108... + 200)Required Addition = S' - S''We know that, S_n = n/2 (a + l):. S_101 = 101/2 * (100 + 200)= 101/2 (300)= 15150Numbers between 100 and 200 divisible by 4 are 100, 104, 108 ...Which are in arithmetic progression.In which a=100 and d=4Let n be the term such that f(n) = 200We know that, f(n) = a + (n - 1)d100 + (n - 1)(4) = 200(n - 1)(4) = 100n - 1 = 25n = 26We know that, S_n = n/2 [2a + (n - 1)d]:. S_26 = 26/2 * [2(100) + (26 - 1)(4)]= 13 * [200 + (100)]= 13 * [300]= 3900:. Required Addition = 15150 - 3900= 11250

 Problem : 14 [ Arithmetic Progression ]       Solve this type of problem 14. For arithemetic progression addition of three terms is 15 and addition of their squres is 83 , then find that nos. Show Answer A: Let the terms are a-d, a, a+dAddition of this terms is 15=> (a-d) + a + (a+d) = 15=> 3 a = 15=> a = 15/3 = 5Addition of their square is 83=> (a-d)^2 + a^2 + (a+d)^2 = 83=> a^2 - 2ad + d^2 + a^2 + a^2 + 2ad + d^2 = 83=> 3a^2 + 2d^2 = 83=> 3(5)^2 + 2d^2 = 83=> 2d^2 = 8=> d^2 = 4=> d = +- 2d = +2 => Required terms : 5 - 2, 5, 5 + 2 => 3, 5, 7d = -2 => Required terms : 5-(-2), 5, 5-2 => 7, 5, 3

 Problem : 15 [ Arithmetic Progression ]       Solve this type of problem 15. If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2 - S1) Show Answer A: Let a be the first term and d be the common differenceNow,S_1= n/2 [ 2a + (n - 1) d ]S_2 = (2n)/2 [ 2a + (2n - 1) d ]S_3 = (3n)/2 [ 2a + (3n - 1) d ]Now, 3(S_2 - S_1)= 3 [ (2n)/2 ( 2a + (2n - 1) d ) - n/2 ( 2a + (n - 1) d ) ]= 3 [ n/2 ( 2(2a) - 2a ) + n/2 ( 2(2n - 1) d - (n - 1) d ) ]= 3 [ n/2 ( 2a ) + n/2 ( 4n - 2 - n + 1) d ) ]= 3 [ n/2 ( 2a ) + n/2 ( 3n - 1) d ) ]= (3n)/2 [ 2a + ( 3n - 1) d ]= S_3 (Proved)

 Problem : 16 [ Arithmetic Progression ]       Solve this type of problem 16. If Sn is sum of n even terms of arithmetic progression series and Sn' is sum of n odd terms of arithmetic progression series then prove that Sn = (1 + 1/n) Sn' Show Answer A: Here S_n = 2 + 4 + 6 + ... + 2n:. S_n = n/2 [ 2a + (n - 1) d ]= n/2 [ 2(2) + (n - 1) * 2 ] (because a = 2 and d = 2)= n/2 [ 4 + 2n - 2 ]= n/2 [ 2n + 2 ]= n ( n + 1 ) Now, S_(n') = 1 + 3 + 5 + ... + (2n - 1):. S_(n') = n/2 [ 2a + (n - 1) d ]= n/2 [ 2(1) + (n - 1) * 2 ] (because a = 1 and d = 2)= n/2 [ 2 + 2n - 2 ]= n/2 [ 2n ]= n^2Now, (S_n)/(S_n') = (n (n + 1))/(n^2) :. (S_n)/(S_n') = ((n + 1))/n:. S_n = (1 + 1/n) × S_(n') (Proved)

 Problem : 17 [ Arithmetic Progression ]       Solve this type of problem 17. For arithemetic progression, addition of three terms is 51 and multiplication of end terms is 273 , then find that nos. Show Answer A: Let the terms are a-d, a, a+dAddition of this terms is 51=> (a-d) + a + (a+d) = 51=> 3 a = 51=> a = 51/3 = 17Multiplication of last 2 terms is 273=> (a-d)(a+d) = 273=> a^2 - d^2 = 273=> d^2 = a^2 - 273=> d^2 = (17)^2 - 273=> d^2 = 289 - 273=> d^2 = 16=> d = +- 4d = +4 => Required terms : 17 - 4, 17, 17 + 4 => 13, 17, 21d = -4 => Required terms : 17-(-4), 17, 17-4 => 21, 17, 13

 Problem : 18 [ Arithmetic Progression ]       Solve this type of problem 18. For arithemetic progression of four terms, addition of end terms is 14 and multiplication of middle two terms is 45 , then find that nos. Show Answer A: Let the terms are a-3d, a-d, a+d, a+3dAddition of last 2 term is 14=> (a-3d) + (a+3d) = 14=> 2a = 14=> a = 14/2 = 7Now multiplication of middle two terms is 45=> (a-d) (a+d) = 45=> a^2 - d^2 = 45=> d^2 = a^2 - 45=> d^2 = 7^2 - 45=> d^2 = 49 - 45=> d^2 = 4=> d = +- 2d = +2 => Required terms : 7-3(2), 7-2, 7+2, 7+3(2) => 1, 5, 9, 13d = -2 => Required terms : 7-3(-2), 7-(-2), 7-2, 7-3(2) => 13, 9, 5, 1

 Problem : 19 [ Arithmetic Progression ]       Solve this type of problem 19. For arithemetic progression, addition of 4 terms is 4 and addition of multiplication of end terms and multiplication of middle terms is -38 , then find that nos. Show Answer A: Let the terms are a-3d, a-d, a+d, a+3dAddition of this terms is 4=> (a-3d) + (a-d) + (a+d) + (a+3d) = 4=> 4 a = 4=> a = 4/4 = 1Addition of multiplication of last 2 terms and middle 2 terms is -38=> (a-3d) (a+3d) + (a-d) (a+d) = -38=> (a^2 - 9d^2) + (a^2 - d^2) = -38=> 2a^2 - 10d^2 = -38=> 2(1)^2 - 10d^2 = -38=> 2 - 10d^2 = -38=> 10d^2 = 40=> d^2 = 4=> d = +- 2d = +2 => Required terms : 1-3(2), 1-2, 1+2, 1+3(2) => -5, -1, 3, 7d = -2 => Required terms : 1-3(-2), 1-(-2), 1-2, 1-3(2) => 7, 3, -1, -5