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Problem 19 of 19 

19. For arithemetic progression, addition of 4 terms is and addition of multiplication of end terms and multiplication of middle terms is , then find that nos.










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Arithmetic Progression 
This is sample question and answer for help purpose.
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.
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`

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`










