The ratio of the sum of m and n terms of an A.P. is `m^(2)` :`n^(2)`. Show that the ratio `m^(th)` and `n^(th)` term is `(2m-1) : (2n-1).`

Answer

N/a

Explanation

Let first term be 'a' and common difference be 'd' of A.P.
`"Now, "(S_(m))/(S_(n))=(m^(2))/(n^(2))`
`rArr" "((m)/(2)[2a+(m-1)d])/((n)/(2)[2a+(n-1)d])=(m^(2))/(n^(2))`
`rArr" "(2a+(m-1)d)/(2a+(n-1)d)=(m)/(n)`
`2an+(mn-n)d=2am+(mn-m)d`
`2a(n-m)=d(n-m)`
`rArr" "2a=d`
`"Now "(T_(m))/(T_(n))=(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).`