The ratio of the sums of m and n terms of an A.P. is m^2: n^2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1).

Question

hartnn · Answer

u need to do some algebra here ....
i assume you know the formulas foe sum and n'th term of AP
did u reach to a point where u get 2a=d ??

hartnn · Answer

write formulas for sum , cancel m and n , cross-multiply, u get 
(2a-d)n=(2a-d)m
means 2a=d
now write the n'th term formual
put d=2a and cancel a from numerstor and denom., u'll get you result.

hartnn · Answer

sorry for typos.....

anonymous · Answer

Well, that also means m=n .. And thus, ratio of two same things is 1 .. No ?

hartnn · Answer

No.

hartnn · Answer

if (2a-d)n =(2a-d)m 
since m and n are distinct, 2a-d = 0 will always satisfy (2a-d)n =(2a-d)m
so, 2a=d

hartnn · Answer

u can look it this way also
2an-nd = 2am-md
2a(n-m)=d(n-m)
so dividing by m-n as m-n$\ne$0
we get 2a=d

hartnn · Answer

still not convinced ?

anonymous · Answer

Sort of! :P

hartnn · Answer

did u get the ratio as  (2m – 1): (2n – 1)
and explicitly mention where u have doubt ?

anonymous · Answer

Nai nai nai, I AM convinced :D
THANK YOU! :P :D

hartnn · Answer

welcome ^_^