The ratio of the sums of m and n terms of an A.P. is m 2 : n 2 . Show that the ratio of m th and n th term is (2 m – 1): (2 n – 1).

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Solution

Let a be the first term of the A.P. and d be the common difference.

It is given that the ratio of the sum of m and n terms of an A.P. is m 2 : n 2 . So,

sum of m terms sum of n terms = m 2 n 2 m 2 [ 2a+( m−1 )d ] n 2 [ 2a+( n−1 )d ] = m 2 n 2 2a+( m−1 )d 2a+( n−1 )d = m n (1)

Substitute m=2m−1 and n=2n−1 in equation (1),

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

It is known that n th term of an A.P. is a+( n−1 )d. So,

m th term of an A.P. n th term of an A.P. = 2m−1 2n−1

Hence, it is proved that the ratio of m th term and n th term of an A.P. is ( 2m−1 ):( 2n−1 ).

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