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Permutations And Combinations

Question
CBSEENMA11013129

The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is 87 1 half. Find them.

Solution

Let the three numbers be space space straight a over straight r comma space straight a comma space ar
Since their product is 125
∴             WiredFaculty   rightwards double arrow   WiredFaculty   rightwards double arrow  a = 5
Since the sum of their products taken in pairs is 87 1 half
∴   space space open parentheses straight a over straight r close parentheses left parenthesis straight a right parenthesis plus left parenthesis straight a right parenthesis space left parenthesis ar right parenthesis space plus space left parenthesis ar right parenthesis open parentheses straight a over straight r close parentheses space equals space 87 1 half    rightwards double arrow   straight a squared over straight r plus straight a squared straight r plus straight a squared space equals space 175 over 2
rightwards double arrow   space space space space space space space space 25 over straight r plus 25 straight r plus 25 space equals space 175 over 2               rightwards double arrow    25 plus 25 straight r squared plus 25 straight r space equals space 175 over 2 straight r
rightwards double arrow          50 straight r squared minus 125 straight r plus 50 equals 0                 rightwards double arrow        2 straight r squared minus 5 straight r plus 2 equals 0

rightwards double arrow                              straight r equals fraction numerator 5 plus-or-minus square root of 25 minus 16 end root over denominator 4 end fraction space equals space fraction numerator 5 plus-or-minus 3 over denominator 4 end fraction equals 2 comma space 1 half
 When a = 5 and r = 2,  numbers are 5 over 2 comma space space 5 comma space space 5 left parenthesis 2 right parenthesis  or  WiredFaculty
When a = 5 and r = 1 half comma  numbers are fraction numerator 5 over denominator begin display style 1 half end style end fraction comma space 5 comma space 5 open parentheses 1 half close parentheses  or 10, 5, 5 over 2

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.