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If a, b, c are in GP, prove that `1/((a+b)), 1/(2b), 1/(b+c)` are in AP. |
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Answer» Let `b=ar` and `c=ar^(2)`. Then, `1/(a+b)=1/(a+ar)=1/(a(1+r)), 1/(2b)=1/(2ar), and 1/((b+c))=1/(ar(1+r))` `:. 1/((a+b))+1/((b+c))=1/(a(1+r))=1/(ar(1+r))=((1+r))/(ar(1+r))=1/(ar)=2xx(1/(2b))`. Hence, `1/((a+b)), 1/(2b), 1/((b+c))` are in AP. |
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