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In a set of four numbers, the first thre...

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 of the set which are in A.P. be b, b + 6, b + 12 and the first number be a
Thus , the four numbers are a,b,b+ 6, b+12
Given , a = b + 12 …(i)
Also a, b,b + 6 are in G.P.
`rArr b + 12 , b, + 6 ` are in G.P. [from (i)]
`therefore (b)/(b+12) = (b+6)/(b)`.
`rArr b^(2) =(b+ 6 ) (b + 12) " "rArr b^(2) = b^(2) + 18b + 72 `
Also, 18b = - 72
`rArr b = - 4 ` " " `rArr a = - 4 + 12 = 8 `
Hence the four numbers are 8, -4 , 2 and 8 .
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