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In a set of four numbers, the first three are in GP and the last three are in AP with a common difference of 6. If the number is same as the fourth, then find the four numbers.

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Let the last three numbers in AP, be `a,a+6,a+12` .
`" " " " [:." 6 is the common difference "]`
If first number is b, then four numbes are
`b,a,a+6,a+12`
But given, `b=a+12`
`:.` Four numbers are `a+12,a,a+6,a+12" " ".....(i)"`
Since, first three numbers are in GP.
Then, `a^(2)=(a+12)(a+6)`
`implies a^(2)=a^(2)+18a+72`
`implies 18a+71=0`
`:. " " a=-4 " " [" from Eq.(i) "]`
Hence, four numbers are `8,-4,2,8` .
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ARIHANT MATHS-SEQUENCES AND SERIES-Exercise (Questions Asked In Previous 13 Years Exam)
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  12. in a geometric progression consisting of positive terms, each term eq...

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