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For arithmetic progression Sm = n and Sn = m then prove that Sm+n = -(m - n)

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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)






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