document.write( "Question 1151136: Three numbers are in an arithmetic progression with a common difference of 6. If 4 is subtracted from the first number, 1 is subtracted from the second number, and the third number is first decreased by 3 and then multiplied by 3, then the resulting three numbers form a geometric progression. Find the original three numbers. \n" ); document.write( "
Algebra.Com's Answer #772771 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "Let the original three numbers be x-6, x, and x+6.

\n" ); document.write( "Then (x-6)-4 = x-10, x-1, and 3((x+6)-3) = 3x+9 form a geometric progression, which means there is a common ratio between the terms.

\n" ); document.write( "\"%28x-1%29%2F%28x-10%29+=+%283x%2B9%29%2F%28x-1%29\"
\n" ); document.write( "\"3x%5E2-21x-90+=+x%5E2-2x%2B1\"
\n" ); document.write( "\"2x%5E2-19x-91+=+0\"
\n" ); document.write( "\"%282x%2B7%29%28x-13%29+=+0\"

\n" ); document.write( "\"x+=+-7%2F2\" or \"x+=+13\"

\n" ); document.write( "Both solutions satisfy the conditions of the problem.

\n" ); document.write( "(A) x = -7/2:

\n" ); document.write( "The original arithmetic progression is

\n" ); document.write( "-19/2, -7/2, 5/2

\n" ); document.write( "When 4 is subtracted from the first term, 1 is subtracted from the second term, and the third term is decreased by 3 and then multiplied by 3, the resulting numbers form a geometric progression:

\n" ); document.write( "-27/2, -9/2, -3/2

\n" ); document.write( "(B) x = 13:

\n" ); document.write( "The original arithmetic progression is

\n" ); document.write( "7, 13, 19

\n" ); document.write( "When 4 is subtracted from the first term, 1 is subtracted from the second term, and the third term is decreased by 3 and then multiplied by 3, the resulting numbers form a geometric progression:

\n" ); document.write( "3, 12, 48

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