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Problem 1 of 19 

1. For given arithemetic progression series ,... find th term and addition of first th terms.










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Arithmetic Progression 
This is sample question and answer for help purpose.
1. For given arithemetic progression series 7,3,1,5,9 ,... find 10 th term and addition of first 10 th terms.
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 : 1 / 19
[ Arithmetic Progression ]
1.
For given arithemetic progression series 7,3,1,5,9 ,... find 10 th term and addition of first 10 th terms.




Solution:

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 / 19
[ Arithmetic Progression ]
2.
For arithemetic progression f( 5 ) = 56 , f( 8 ) = 86 then find f( 10 ) and S( 10 ).




Solution:

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 / 19
[ Arithmetic Progression ]
3.
For arithemetic progression f( 5 ) = 25 , f( 11 ) = 49 , then find n such that f(n) = 105 .




Solution:

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 / 19
[ Arithmetic Progression ]
4.
For arithemetic progression S( 33 ) = 198 , then find f( 17 ).




Solution:

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 / 19
[ Arithmetic Progression ]
5.
For arithemetic progression f( 17 ) = 6 , then find S( 33 ).




Solution:

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 / 19
[ Arithmetic Progression ]
6.
For arithemetic progression f( 7 ) = 13 , S( 14 ) = 203 , then find f( 10 ) and S( 8 ).




Solution:

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 / 19
[ Arithmetic Progression ]
7.
For arithemetic progression addition of 3 terms is 27 and their multiplication is 648 , then that nos.




Solution:

Let the terms are `ad, a, a+d`
Addition of this terms is `27`
`=> (ad) + a + (a+d) = 27`
`=> 3 a = 27`
`=> a = 27/3 = 9`
Multiplication of this terms is `648`
`=> (ad) × a × (a+d) = 648`
`=> (9d) × 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, 93 => 12, 9, 6`





Problem : 8 / 19
[ Arithmetic Progression ]
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.




Solution:

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 / 19
[ Arithmetic Progression ]
9.
For arithmetic progression Sm = n and Sn = m then prove that Sm+n = (m  n)




Solution:

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 / 19
[ Arithmetic Progression ]
10.
For arithmetic progression Sm = n and Sn = m then prove that Smn = (m  n)(1 + 2n / m)




Solution:

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_(mn) = (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 / 19
[ Arithmetic Progression ]
11.
The ratio of two arithemetic progression series is 3x+5 : 4x2 , then find the ratio of their 10 th term.




Solution:

Let two series be `a, a+d, a+2d, ...` and `A, A+D, A+2D, ...`
Now, `S_n / S_(n') = (3x+5) / (4x2)`
`=> (n/2 [2a + (n  1)d]) / (n/2 [2A + (n  1)D]) = (3x+5) / (4x2)`
`=> [2a + (n  1)d] / [2A + (n  1)D] = (3x+5) / (4x2) >(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) / (4x2),` where `n=19`
`=> f(10)/(F(10)) = 62 / 74`
`=> f(10)/(F(10)) = 31/37`





Problem : 12 / 19
[ Arithmetic Progression ]
12.
Find the sum of all natural nos between 100 to 200 and which are divisible by 4 .




Solution:

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 / 19
[ Arithmetic Progression ]
13.
Find the sum of all natural nos between 100 to 200 and which are not divisible by 4 .




Solution:

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 / 19
[ Arithmetic Progression ]
14.
For arithemetic progression addition of three terms is 15 and addition of their squres is 83 , then find that nos.




Solution:

Let the terms are `ad, a, a+d`
Addition of this terms is `15`
`=> (ad) + a + (a+d) = 15`
`=> 3 a = 15`
`=> a = 15/3 = 5`
Addition of their square is `83`
`=> (ad)^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, 52 => 7, 5, 3`





Problem : 15 / 19
[ Arithmetic Progression ]
15.
If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2  S1)




Solution:

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 / 19
[ Arithmetic Progression ]
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'




Solution:

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 / 19
[ Arithmetic Progression ]
17.
For arithemetic progression, addition of three terms is 51 and multiplication of end terms is 273 , then find that nos.




Solution:

Let the terms are `ad, a, a+d`
Addition of this terms is `51`
`=> (ad) + a + (a+d) = 51`
`=> 3 a = 51`
`=> a = 51/3 = 17`
Multiplication of last 2 terms is `273`
`=> (ad)(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, 174 => 21, 17, 13`





Problem : 18 / 19
[ Arithmetic Progression ]
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.




Solution:

Let the terms are `a3d, ad, a+d, a+3d`
Addition of last 2 term is `14`
`=> (a3d) + (a+3d) = 14`
`=> 2a = 14`
`=> a = 14/2 = 7`
Now multiplication of middle two terms is `45`
`=> (ad) (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 : `73(2), 72, 7+2, 7+3(2) => 1, 5, 9, 13`
`d = 2 =>` Required terms : `73(2), 7(2), 72, 73(2) => 13, 9, 5, 1`





Problem : 19 / 19
[ Arithmetic Progression ]
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.




Solution:

Let the terms are `a3d, ad, a+d, a+3d`
Addition of this terms is `4`
`=> (a3d) + (ad) + (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`
`=> (a3d) (a+3d) + (ad) (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 : `13(2), 12, 1+2, 1+3(2) => 5, 1, 3, 7`
`d = 2 =>` Required terms : `13(2), 1(2), 12, 13(2) => 7, 3, 1, 5`







