Home

Solve any problem
(step by step solutions)
Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution
For arithmetic progression Sm = n and Sn = m then prove that Sm+n = -(m - n)

Solution:
Your problem `->` For arithmetic progression Sm = n and Sn = m then prove that Sm+n = -(m - n)


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)






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