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The ratio of the sum of `ma n dn` terms of an A.P. is `m^2: n^2dot` Show that the ratio of the mth and nth terms is `(2m-1):(2n-1)dot`
A. `(2m+1):(2n-1),`
B. `m:n`
C. `(2m-1):(2n-1)`
D. None of these

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(c) Here, `" "(S_(m))/S_(n)=(m^2)/(n^2) " "[therefore A=1,B=0]` S
` therefore " " (t_(m))/(t_(n))=((2m-1))/((2n-1))`
`implies t_(m):t_(n)= (2m-1):(2n-1)`

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