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Question
If tk is the kth term of a G.P., then show that tn – k, tn, tn + k also form a GP for any positive integer k
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Solution
Let a be the first term and r be the common ratio.
We are given tk = ark – 1
We have to prove: tn – k, tn, tn + k form a G.P.
tn – k = `"ar"^("n" - "k" - 1)`
tn = arn – 1
tn + k = `"ar"^("n" + "k" - 1)`
Now `"t"_"n"/"t"_("n" - "k") = ("ar"^("n" - 1))/("ar"^("n" - "k" - 1))`
= `"r"^("n"-1 - "n" + "k" + 1)`
= rk
Also `("t"_("n" + "k"))/"t"_"n" =("ar"^("n" + "k" - 1))/("ar"^("n" - 1))`
= `"r"^("" + "k - 1 - "n" + 1)`
= rk
Now `"t"_"n"/("t"_("n" -"k")) = ("t"_("n"+ "k"))/"t"_"n"`
⇒ tn – k, tn, tn + k fom a G.P.
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