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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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Answer» We have, `2q=p+r`. Let `T_(p), T_(q), T_(r)` be the given terms of a GP with terms =A common ratio = R. Then, `(T_(q))^(2)={AR^((q-1))}^(2)=A^(2)R^((2q-2))` and `(T_(p)xxT_(r))= {AR^((p-1))xxAR^((r-1))}=A^(2)R^((p+r-2)=A^(2) R^((2q-2))`. `:. (T_(p))^(2)=(T_(p)xxT_(r))` and hence `T_(p), T_(q), T_(r)` are in GP. |
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