Question

The ratio of the sums of first m and first n terms of an A. P. is m² : n² Show that the ratio of its mth and nth terms is (2m-1):(2n-1).

Answer 1

Given that : The ratio of the sums of first m and n terms of an A.P. is m²:n²

Then, we have to prove that :

The ratio of its mth and nth term is (2m-1):(2n-1)

Now, according to the question :

$\frac{ \frac{m}{2}[2a +(m - 1)d ]}{ \frac{n}{2}[2a + (n - 1)d ]} = \frac{ {m}^{2} }{ {n}^{2} } \\ \\ = > \frac{[2a + (m - 1)d]}{[2a + (n - 1)d]} = \frac{m}{n} \\ \\ = > \frac{(2a + md - d)}{(2a + nd - d)} = \frac{m}{n} \\ \\ = > 2an + mnd - dn = 2am + mnd - md \\ \\ = > 2an - dn = 2am - md \\ \\ = > 2an - 2am = dn - dm \\ \\ = > 2a(n - m) = d(n - m) \\ \\ = > 2a = d$

Now, ratio of mth term and nth term of A.P.

$\frac{a + (m - 1)d}{a + (n - 1)d} \\ \\ = > \frac{a + (m - 1)2a}{a + (n - 1)2a } ( d=2a ) \\ \\ = > \frac{a + 2am - 2a}{a + 2an - 2a} \\ \\ = > \frac{2am - a}{2an - a} \\ \\ = > \frac{a(2m - 1)}{a(2n - 1)} \\ \\ = > \frac{(2m - 1)}{(2n - 1)} \\ \\ = > (2m - 1):(2n - 1)$

Then the ratio of mth term and nth term of A.P. will be (2m-1):(2n-1)

HENCE PROVED

Answer 2

Solution :

Let a be the first term and d be the

common difference of the given A.P.

The sum of first m and n terms are

given by

Sm = (m/2)[ 2a + ( m - 1 )d ] ,

Sn = (n/2)[ 2a + ( n -1 )d ]

Sm /Sn = m²/n²

=>{(m/2)[2a+(m-1)d]}/{(n/2)[2a+(n-1)d]}=m²/n²

=> [2a+(m-1)d]/[2a+(n-1)d] = m/n

=> [2a+(m-1)d ]n = [ 2a+(n-1)d ]m

=> 2an + (m-1)nd = 2am + (n-1)md

=> 2an - 2am = [ (n-1)m - (m-1)n ]d

=> 2a( n - m ) = [ mn - m - mn + n ]d

=> 2a( n - m ) = ( n - m )d

=> 2a = d ----( 1 )

Now ,

Ratio of mth and nth terms = tm/tn

= [ a + ( m - 1 )d ]/[ a + ( n - 1 )d ]

= [a + (m - 1)2a]/[a + (n - 1)2a] { from ( 1 ) }

= [a( 1 + 2m - 2 )]/[a( 1 + 2n - 2 )]

= ( 2m - 1 )/( 2n - 1 )

•••••