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If the `pth , qth` and `rth`terms of aG.P. are `a , b , c`respectively, prove that: `a^(q-r).b^(r-p)c^(p-q)=1.` |
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Answer» Let A be the first term and R be the common ratio of the given GP. Then, `T_(p)=a rArr a=AR^((p-1))` `T_(q)=b rArr b=AR^((q-1))` `T_(r)=c rArr c= AR^((r-1))`. `:. a^((q-r)).b^((r-p)).c^((p-q))` `={AR^((p-1))}^((q-r)).{AR^((q-1))}^((r-p)).{AR^((r-1))}^((p-q))` `=A^((q-r)).R^((p-1)(q-r)).A^((r-p)).R^((q-1)(r-p)).A^((p-q)).R^((r-1)(p-q))` `=A^({(q-r)+(r-p)+(p-q)}).R^({(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)})` `=A^(0).R^(p(q-r)+q(r-p)+r(p-q)+(r-q)+(p-r)+(q-p))` `=(1xxR^(0))=(1xx1)=1`. Hence, `a^((q-r)).b^((r-p)).c^((p-q))=1`. |
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