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Question
If the ratio of sum of the first m and n terms of an AP is m2 : n2, show that the ratio of its mth and nth terms is (2m − 1) : (2n − 1) ?
Solution
Let the first term and the common difference of the AP be a and d, respectively.
Therefore,
Sum of the first m terms of the AP,
`"S"_"m" = "m"/2 [2"a" + ("m - 1")"d"]`
Sum of the first n terms of the AP,
`"S"_"n" = "n"/2[2"a" + ("n - 1")"d"]`
It is given that
`"S"_"m"/"S"_"n" = ("m"/2 [2"a" + ("m - 1")"d"])/("n"/2[2"a" + ("n - 1")"d"])`
`=> ([2"a" + ("m - 1")"d"])/([2"a" + ("n - 1")"d"]) = "m"/"n"`
⇒ 2an + mnd - nd = 2am + nmd - md
⇒ 2an - 2am = nd - md
⇒ 2a(n - m) = d(n - m)
⇒ 2a = d
Now,
`"T"_"m"/"T"_"n" = ("a" + ("m" - 1)"d")/("a" + ("n" - 1)"d")`
`=> "T"_"m"/"T"_"n" = ("a" + ("m - 1") xx 2"a")/("a" + ("n" - 1) xx 2"a")`
`=> "T"_"m"/"T"_"n" = (2"m" - 1)/("2n" - 1)`
Hence, the ratio of the mth term to the nth term is (2m − 1) : (2n − 1).
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