Show that the sum of ( m + n ) th and ( m – n ) th terms of an A.P. is equal to twice the m t h term.
Let, a be the first term and d be the common difference of an A.P.
The ( m + n ) t h term of an A.P. is,
a m + n = a + ( m + n − 1 ) d
The ( m − n ) t h term of an A.P. is,
a m − n = a + ( m − n − 1 ) d
The m t h term of an A.P. is,
a m = a + ( m − 1 ) d
Add ( m + n ) t h term and ( m − n ) t h term.
a m + n + a m − n = a + ( m + n − 1 ) d + a + ( m − n − 1 ) d = 2 a + ( m + n − 1 + m − n − 1 ) d = 2 a + ( 2 m − 2 ) d = 2 [ a + ( m − 1 ) d ]
This shows that the sum of both the terms is equal to 2 a m .
Hence, it is proved that the sum of ( m + n ) t h term and ( m − n ) t h term is equal to twice of m t h term.
Let,
The
The
The
Add
This shows that the sum of both the terms is equal to
Hence, it is proved that the sum of
