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If `p, q, r` are in AP then prove that pth, qth and rth terms of any GP are in GP.

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Since, p, q, r are in AP, we have `(q-p)=(r-q)`
Let us consider a GP with first term `=A` and common ratio `=R`.
Then, `T_(p)=AR^((p-1)), T_(q)=AR^((q-1))` and `T_(r)=AR^((r-1))`.
`:. T_(q)/T_(p)=(AR^((q-1)))/(AR^((p-1)))=R^((q-p))=R^((r-q))` [using (i)]
and `T_(r)/T_(q)=(AR^((r-1)))/(AR^((q-1)))=R^((r-q))`.
Thus, `T_(r)/T_(p)=T_(r)/T_(q)` and therefore, `T_(p), T_(q), T_(r)` are in GP.
Hence, the pth, qth and rth terms of any GP are in GP.

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