1.

ththe same as the Sum otiS Zer0

Answer»

m/2 * (2a + (m-1)d) = n/2 * (2a + (n-1)d)

we get

m(2a + (m-1)d) - n(2a+ (n-1)d) = 0

2am + m^2d - md -2an -n^2d +nd =0

2a(m-n) + (m^2 - n^2)d -(m - n)d =0

2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0

taking (m-n) common

2a + ( m + n -1) d = 0------------ 1

S m+n = m+n/2( 2a + (m+n -1)d

we know that 2a + (m+n)d is 0 from eqn. 1

therefore S m+n = 0


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