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If a, b, c are in AP and a, b, d are in GP, show that `a, (a-b)` and `(d-c)` are in GP. |
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Answer» `(a+c)=2b and b^(2)=ad`. We have to prove that `(a-b)^(2)=a(d-c)`. Now, `(a-b)^(2)=a^(2)+b^(2)-2ab=a^(2)+ad-a(a+c)` `rArr (a-b)^(2)=ad-ac=a(d-c)`. Hence, `a(a-b), (d-c)` are in GP. |
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