Question

this is challenge for all.
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.

Answer 1

$Hey\:there$‼

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

$Now$, 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 )