Home

Solve any problem
(step by step solutions)
Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution
If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2 - S1)

Solution:
Your problem `->` If S1, S2, S3 are sum of n, 2n, 3n terms of arithmetic progression series then prove that S3 = 3(S2 - S1)


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)






Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
Want to know about AtoZmath.com and me
  
 

Share with your friends, if solutions are helpful to you.
 
Copyright © 2019. All rights reserved. Terms, Privacy