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 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

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