Permutations And Combinations

Question
CBSEENMA11013137

If pth, qth, rth and sth terms of an A.P. be in G.P., then prove that (p – q), (q – r), (r – s) are in GP.

Solution

Let A be the first term and D be the common difference.
∴                     straight t subscript straight p space equals space straight A plus left parenthesis straight p minus 1 right parenthesis straight D comma space space space space straight t subscript straight q space equals space straight A plus left parenthesis straight q minus 1 right parenthesis straight D comma space space straight t subscript straight r space equals space straight A plus left parenthesis straight r minus 1 right parenthesis straight D comma
                      straight t subscript straight s space equals space straight A plus left parenthesis straight s minus 1 right parenthesis straight D
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#6 {main}</pre>                                                              ...(i)
                 straight t subscript straight q minus straight t subscript straight r space equals space left parenthesis straight q minus straight r right parenthesis straight D                                                              ...(ii)
                straight t subscript straight r minus straight t subscript straight s space equals space left parenthesis straight r minus straight s right parenthesis straight D                                                                ...(iii)
Since straight t subscript straight p comma space straight t subscript straight q comma space straight t subscript straight r comma space straight t subscript straight s are in G.P.
∴                         straight t subscript straight q over straight t subscript straight p space equals space straight t subscript straight r over straight t subscript straight q space equals space straight t subscript straight s over straight t subscript straight r space equals space space straight R left parenthesis say right parenthesis comma space space straight t subscript straight q equals straight t subscript straight p straight R comma space straight t subscript straight r space equals space straight t subscript straight q straight R space equals straight t subscript straight p straight R squared comma space straight t subscript straight s equals straight t subscript straight r straight R space equals space straight t subscript straight p straight R cubed
Now,                      space space straight t subscript straight q over straight t subscript straight p equals space straight t subscript straight r over straight t subscript straight q  and straight t subscript straight r over straight t subscript straight q space equals space straight t subscript straight s over straight t subscript straight r
rightwards double arrow         straight t subscript straight q over straight t subscript straight p minus 1 space equals space straight t subscript straight r over straight t subscript straight q minus 1  and straight t subscript straight r over straight t subscript straight q minus 1 space equals space straight t subscript straight s over straight t subscript straight r minus 1
rightwards double arrow         fraction numerator straight t subscript straight q minus straight t subscript straight p over denominator straight t subscript straight p end fraction space equals space fraction numerator straight t subscript straight r minus straight t subscript straight q over denominator straight t subscript straight p straight R end fraction  and  space space fraction numerator straight t subscript straight r minus straight t subscript straight q over denominator straight t subscript straight p straight R end fraction space equals space fraction numerator straight t subscript straight s minus straight t subscript straight r over denominator straight t subscript straight p straight R squared end fraction
rightwards double arrow         fraction numerator negative left parenthesis straight p minus straight q right parenthesis straight D over denominator 1 end fraction space equals space fraction numerator negative left parenthesis straight q minus straight r right parenthesis straight D over denominator straight R end fraction and fraction numerator negative left parenthesis straight q minus straight r right parenthesis straight D over denominator 1 end fraction space equals space fraction numerator negative left parenthesis straight r minus straight s right parenthesis straight D over denominator straight R end fraction
rightwards double arrow         fraction numerator straight q minus straight r over denominator straight p minus straight q end fraction space equals space straight R and fraction numerator straight r minus straight s over denominator straight q minus straight r end fraction space equals space straight R
rightwards double arrow          space space fraction numerator straight q minus straight r over denominator straight p minus straight q end fraction space equals space fraction numerator straight r minus straight s over denominator straight q minus straight r end fraction
rightwards double arrow      p - q, q - r, r - s are in G.P.

Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.