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For arithmetic progression Sm = n and Sn = m then prove that Sm-n = (m - n)(1 + 2n / m) [ Calculator, Method and examples ]

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Your problem `->` For arithmetic progression Sm = n and Sn = m then prove that Sm-n = (m - n)(1 + 2n / m)


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)






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