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 = -4` We know that, `f(n) = a + (n - 1)d` `f(10) = 7 + (10 - 1)(-4)` `= 7 + (-36)` `= -29` We know that, `S_n = n/2 [2a + (n - 1)d]` `:. S_10 = 10/2 * [2(7) + (10 - 1)(-4)]` `= 5 * [14 + (-36)]` `= 5 * [-22]` `= -110` Hence, `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)d` `f(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 = 10` We know that, `f(n) = a + (n - 1)d` `f(10) = 16 + (10 - 1)(10)` `= 16 + (90)` `= 106` We 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)d` `f(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 = 4` Let `n` be the term such that `f(n) = 105` We know that, `f(n) = a + (n - 1)d` `9 + (n - 1)(4) = 105` `(n - 1)(4) = 96` `n - 1 = 24` `n = 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)d` `f(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)d` `f(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)d` `f(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 = 3` We know that, `f(n) = a + (n - 1)d` `f(10) = -5 + (10 - 1)(3)` `= -5 + (27)` `= 22` We 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+d` Addition of this terms is `27` `=> (a-d) + a + (a+d) = 27` `=> 3 a = 27` `=> a = 27/3 = 9` Multiplication 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 = +- 3` `d = +3 =>` Required terms : `9 - 3, 9, 9 + 3 => 6, 9, 12` `d = -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 = -0.201` From `(1) => 2a + 16d = 2.8235` `=> 2a = 2.8235 - 16d` `=> 2a = 2.8235 - 16 × -0.201` `=> 2a = 2.8235 - -3.2157` `=> 2a = 6.0392` `=> a = 3.0196` We 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)d` `f(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=4` Let `n` be the term such that `f(n) = 200` We know that, `f(n) = a + (n - 1)d` `100 + (n - 1)(4) = 200` `(n - 1)(4) = 100` `n - 1 = 25` `n = 26` We 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)` `= 15150` Numbers between `100` and `200` divisible by `4` are `100, 104, 108 ...` Which are in arithmetic progression. In which `a=100` and `d=4` Let `n` be the term such that `f(n) = 200` We know that, `f(n) = a + (n - 1)d` `100 + (n - 1)(4) = 200` `(n - 1)(4) = 100` `n - 1 = 25` `n = 26` We 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+d` Addition of this terms is `15` `=> (a-d) + a + (a+d) = 15` `=> 3 a = 15` `=> a = 15/3 = 5` Addition 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 = +- 2` `d = +2 =>` Required terms : `5 - 2, 5, 5 + 2 => 3, 5, 7` `d = -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 difference Now, `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^2` Now, `(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+d` Addition of this terms is `51` `=> (a-d) + a + (a+d) = 51` `=> 3 a = 51` `=> a = 51/3 = 17` Multiplication 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 = +- 4` `d = +4 =>` Required terms : `17 - 4, 17, 17 + 4 => 13, 17, 21` `d = -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+3d` Addition of last 2 term is `14` `=> (a-3d) + (a+3d) = 14` `=> 2a = 14` `=> a = 14/2 = 7` Now 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 = +- 2` `d = +2 =>` Required terms : `7-3(2), 7-2, 7+2, 7+3(2) => 1, 5, 9, 13` `d = -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+3d` Addition of this terms is `4` `=> (a-3d) + (a-d) + (a+d) + (a+3d) = 4` `=> 4 a = 4` `=> a = 4/4 = 1` Addition 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 = +- 2` `d = +2 =>` Required terms : `1-3(2), 1-2, 1+2, 1+3(2) => -5, -1, 3, 7` `d = -2 =>` Required terms : `1-3(-2), 1-(-2), 1-2, 1-3(2) => 7, 3, -1, -5`