Question

If the ratio of the sum of the first mth term of an AP is m2:n2.Show that the ratio of its mth and nth term is (2m-1):(2m-1)

Answer 1

hello Frnd

Here is your answer

Given as

pls have a look at the pic


continuation ...

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$= \frac{a + 2am - 2a}{a + 2an - 2a} \\ \frac{2am - a}{2an - a} \\ = \frac{2m - 1}{2n - 1}$

hOpe it helps
jerri

Answer 2

Hey mate..
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Let, the first term and common difference of an AP be a and d respectively.

Given,

Sum of m terms of an A.P. = m/2 [2a + (m -1)d]

Sum of n terms of an A.P. = n/2 [2a + (n -1)d]

A/Q,

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

⇒ [2a + md - d] / [2a + nd - d] = m/n

⇒ 2an + mnd - nd = 2am + mnd - md

⇒ 2an - 2am = nd - md

⇒ 2a (n -m) = d(n - m)

⇒ 2a = d....(1)

Ratio of m th term to n th term:

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

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

= ( a + 2 am - 2a ) / ( a + 2an - 2a )

= ( 2 am - a ) / ( 2 an - a )

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

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

( Hence, Shown )

Hope it helps !!