If the sum of first m terms of an A.R is the same as the sum of its first nterms, show that the sum of its first (m + n) terms is zero.
Anonymous:
i think its A.P. instead of A.R.
Answers
Answered by
9
Solution :
Let a be the first term and d be the
common difference of the given
A.P ,
Sm = Sn
=> (m/2)[2a+(m-1)d ] = (n/2)[ 2a+(n-1)d ]
=> 2am + m²d - md = 2an + n²d - nd
=> 2am-2an + m²d-n²d -md+nd = 0
=> 2a( m - n ) + [(m²-n²)-(m-n)]d = 0
=> ( m-n )[ 2a+(m+n-1 )]d= 0
=> 2a + ( m + n -1 )d = 0 -----( 1 )
{ Since ,m-n≠ 0 }
Now ,
Sm+n = [(m+n)/2]{ 2a+(m+n-1)d }
= ( m+n )/2 × 0 [ from ( 1 ) ]
= 0
•••••
Let a be the first term and d be the
common difference of the given
A.P ,
Sm = Sn
=> (m/2)[2a+(m-1)d ] = (n/2)[ 2a+(n-1)d ]
=> 2am + m²d - md = 2an + n²d - nd
=> 2am-2an + m²d-n²d -md+nd = 0
=> 2a( m - n ) + [(m²-n²)-(m-n)]d = 0
=> ( m-n )[ 2a+(m+n-1 )]d= 0
=> 2a + ( m + n -1 )d = 0 -----( 1 )
{ Since ,m-n≠ 0 }
Now ,
Sm+n = [(m+n)/2]{ 2a+(m+n-1)d }
= ( m+n )/2 × 0 [ from ( 1 ) ]
= 0
•••••
Similar questions