The ratio of the sum of first m terms and first n terms of an ap is m2:n2 show that the ratio of its nth term and its nth term ts (2m - 1 ):(2n - 1 )
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if a be 1st term of ap
and d be common difference
then
Sm=m{2a+(m-1)d}/2
and Sn= n{2a+(n-1)d}/2
or, Sm/Sn=m^2/n^2
by solving them we get d=2a
hence ratio of the mth & nth term=a+(m-1)d/a+(n-1)d=(1+2m-2)/(1+2n-2)=(2m-1)/(2n-1)
proved
and d be common difference
then
Sm=m{2a+(m-1)d}/2
and Sn= n{2a+(n-1)d}/2
or, Sm/Sn=m^2/n^2
by solving them we get d=2a
hence ratio of the mth & nth term=a+(m-1)d/a+(n-1)d=(1+2m-2)/(1+2n-2)=(2m-1)/(2n-1)
proved
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