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| Question 1069603:  Three consecutive terms in an arithmetic sequence is given with a middle one x. The sum of the three terms is 30 and the product of the three terms is 840
 Calculate the three terms
 Answer by ikleyn(52879)
      (Show Source): 
You can put this solution on YOUR website! . Three consecutive terms in an arithmetic sequence is given with a middle one x. The sum of the three terms is 30 and
 the product of the three terms is 840. Calculate the three terms
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The terms are x-d, x and x+d, where d is the common difference.
For x, you have the equation 
(x-d) + x + (x+d) = 30,
which is equivalent to
3x = 30
and gives x =  = 10.
The second equation is 
(x-d)*x*(x+d) = 840,   or
(10 - d)*10*(10 + d) = 840,
which gives  =  ,   or  = 84,   --->  = 100 - 84 = 16,  --->  = 4,  = -4.
The first progression is  {6, 10, 14}.
The second progression is  {14, 10, 6}. There is a bunch of lessons on arithmetic progressions in this site
 - Arithmetic progressions
 - The proofs of the formulas for arithmetic progressions
 - Problems on arithmetic progressions
 - Word problems on arithmetic progressions
 - Chocolate bars and arithmetic progressions
 - One characteristic property of arithmetic progressions
 - Solved problems on arithmetic progressions
 
 
 Also, you have this free of charge online textbook in ALGEBRA-II in this site
 - ALGEBRA-II - YOUR ONLINE TEXTBOOK.
 
 The referred lessons are the part of this online textbook under the topic
 "Arithmetic progressions".
 
 
 
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