Permutations and Combinations

Permutations and Combinations

Question

The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a GP. Find the numbers.

Answer

Let a - d, a, a + d be three numbers in A.P. Since their sum is 21
∴    a - d + a + a + d = 21
rightwards double arrow                        3a = 21    rightwards double arrow a = 7
∴   Numbers are 7 - d, 7, 7 + d
It is given that if the second number is reduced by 1 and third is increased by 1 then resulting three numbers form G.P.
∴  7 - d, 7 - 1, 7 + d + 1 are in G.P. or 7 - d, 6, 8 + d are in G.P.

rightwards double arrow               fraction numerator 6 over denominator 7 minus straight d end fraction space equals space fraction numerator 8 plus straight d over denominator 6 end fraction    rightwards double arrow     (8 + d) (7 - d) = 36
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rightwards double arrow                 d = fraction numerator negative 1 plus-or-minus square root of 1 plus 80 end root over denominator 2 end fraction space equals space fraction numerator negative 1 plus-or-minus 9 over denominator 2 end fraction space equals space 4 comma space minus space 5
When a = 7 and d = 4, numbers are 7 - 4,  7 + 4 or 3, 7, 11
when a = 7 and d = -5, numbers are 7 - (-5), 7, 7 - 5  or 12, 7, 2

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