Question

If Sm= Sn of an AP
Then prove that Sm+n=0



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Answer 1

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Answer 2

Sm = m/2 [ 2a + ( m - 1 ) d]

Sn = n/2 [ 2a + ( n - 1) d]

Sm = Sn

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

m [2a + dm - d] = n[ 2a + dn - d]

2am + dm^2 - dm = 2an + dn ^2 - dn

2am - 2an = dn^2 - dn - dm^2 + dm

2a( m - n ) = d[ n^2 - n - m^2 + m ]

2a( m - n ) = d[ ( n ^2 - m^2) + ( m - n )]

2a( m - n ) = d[ ( n - m ) ( n + m ) + ( m - n)]

2a ( m - n) = d [ {- ( m - n )} ( n + m ) + ( m - n)]

2a ( m - n) = d ( m - n ) [ - ( m + n) + 1 ]

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

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

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

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

S(m+n) = ( m +n ) /2 [ d ( 0 )]

$\fbox{S(m+n)\: =\: 0}$

Hence, PROVED.