Question

the ratio of the sums of first m and first n terms of an AP is m^2:n^2. Show that the ratio of its mth and nth terms is (2m-1):(2n-1).

Answer 1

HELLO DEAR,


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]


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



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]



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



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

SO, THE RATIO OF M TH TERM AND THE N TH TERM OF THE ARITHMETIC SERIES IS (2m - 1) : (2n -1).


I HOPE ITS HELP YOU DEAR,
THANKS