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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 ]Solution: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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