Permutations and Combinations

Permutations and Combinations

Question

If space space left parenthesis straight b minus straight c right parenthesis squared comma space left parenthesis straight c minus straight a right parenthesis squared comma space left parenthesis straight a minus straight b right parenthesis squared are in A.P., prove that fraction numerator 1 over denominator straight b minus straight c end fraction comma space fraction numerator 1 over denominator straight c minus straight a end fraction comma space fraction numerator 1 over denominator straight a minus straight b end fraction are in A.P.

Answer

Since space left parenthesis straight b minus straight c right parenthesis squared comma space left parenthesis straight c minus straight a right parenthesis squared comma space left parenthesis straight a minus straight b right parenthesis squared are in A.P.

∴  space space space left parenthesis straight c minus straight a right parenthesis squared minus space left parenthesis straight b minus straight c right parenthesis squared space equals space left parenthesis straight a minus straight b right parenthesis squared minus left parenthesis straight c minus straight a right parenthesis squared
or    left parenthesis straight c minus straight a plus straight b minus straight c right parenthesis left parenthesis straight c minus straight a minus straight b plus straight c right parenthesis space equals space left parenthesis straight a minus straight b plus straight c minus straight a right parenthesis left parenthesis straight a minus straight b minus straight c plus straight a right parenthesis
or     (b - a) (2c - a - b) = (c - b) (2a - b - c)                                  ...(i)
Now, fraction numerator 1 over denominator straight b minus straight c end fraction comma space fraction numerator 1 over denominator straight c minus straight a end fraction comma space fraction numerator 1 over denominator straight a minus straight b end fraction will be in A.P.
if         space space space space fraction numerator 1 over denominator straight c minus straight a end fraction minus fraction numerator 1 over denominator straight b minus straight c end fraction space equals space fraction numerator 1 over denominator straight a minus straight b end fraction minus fraction numerator 1 over denominator straight c minus straight a end fraction or if  fraction numerator straight b minus straight c minus straight c plus straight a over denominator left parenthesis straight c minus straight a right parenthesis space left parenthesis straight b minus straight c right parenthesis end fraction = fraction numerator straight c minus straight a minus straight a plus straight b over denominator left parenthesis straight a minus straight b right parenthesis left parenthesis straight c minus straight a right parenthesis end fraction
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or if (b - a) (2c - a - b) = (c - b) (2a - b - c) which is true.                [By using (i)]
Hence, space space fraction numerator 1 over denominator straight b minus straight c end fraction comma space space fraction numerator 1 over denominator straight c minus straight a end fraction comma space fraction numerator 1 over denominator straight a minus straight b end fraction are in A.P.

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