the ratio of the sum of m and n terms of an a.p is msquare : nsquare.prove that the ratio of mth and nth term is 2m-n: 2n-1.
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Given, Ratio of sum of m and terms of an A.P. is m² : n²
Sum of m terms = m / 2 [ 2a + ( m – 1 ) d] ...(1)
Sum of n terms = n / 2 [ 2a + ( n – 1 ) d ] ...(2)
Dividing both (1) and (2) ,
m / 2 [ 2a + ( m – 1 ) d ] . . . m²
_________________ = ___
n / 2 [ 2a + ( n – 1 ) d ] . . . . n²
2a + md – d . . . m
__________ = __
2a + nd – d . . . . n
=> 2an + mnd – nd = 2am + mnd – md
=> 2an – 2am = nd – md
=> 2a ( n – m ) = d ( n – m )
=> 2a = d
, ratio of mth term to nth term :-
[a + ( m – 1 ) d ] / [a + (n – 1 ) d]
=> [ a + ( m – 1 ) 2a ] / [ a + ( n – 1 ) 2a ]
=> a ( 1 + 2m – 2 ) / a ( 1 + 2n – 2 )
=> ( 1 + 2m – 2 ) / ( 1 + 2n – 2 )
=> ( 2m – 1 ) / ( 2n – 1 )
Therefore, ratio of mth term to nth term is ( 2m – 1 ) : ( 2n – 1 )
Given, Ratio of sum of m and terms of an A.P. is m² : n²
Sum of m terms = m / 2 [ 2a + ( m – 1 ) d] ...(1)
Sum of n terms = n / 2 [ 2a + ( n – 1 ) d ] ...(2)
Dividing both (1) and (2) ,
m / 2 [ 2a + ( m – 1 ) d ] . . . m²
_________________ = ___
n / 2 [ 2a + ( n – 1 ) d ] . . . . n²
2a + md – d . . . m
__________ = __
2a + nd – d . . . . n
=> 2an + mnd – nd = 2am + mnd – md
=> 2an – 2am = nd – md
=> 2a ( n – m ) = d ( n – m )
=> 2a = d
, ratio of mth term to nth term :-
[a + ( m – 1 ) d ] / [a + (n – 1 ) d]
=> [ a + ( m – 1 ) 2a ] / [ a + ( n – 1 ) 2a ]
=> a ( 1 + 2m – 2 ) / a ( 1 + 2n – 2 )
=> ( 1 + 2m – 2 ) / ( 1 + 2n – 2 )
=> ( 2m – 1 ) / ( 2n – 1 )
Therefore, ratio of mth term to nth term is ( 2m – 1 ) : ( 2n – 1 )
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