What are the three terms in an A.P. with a sum of 36 and a product of 1428?

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The discussion focuses on finding three consecutive terms in an arithmetic progression (A.P.) that sum to 36 and have a product of 1428. The key equations derived are that the product of the terms is represented as a1 * (a1 + d) * (a1 + 2d) = 1428, where d is the common difference. Additionally, the sum of the terms can be expressed as a1 + (a1 + d) + (a1 + 2d) = 36. By establishing these two equations, the problem can be solved using the variables a1 and d. This approach allows for a systematic way to find the three terms in the A.P.
tykescar
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This isn't a homework question, it's in a textbook I have and I'm a bit stumped. I know there's something relatively simple I'm missing so any help would be much appreciated (working too).

Three consecutive terms of an A.P. have a sum of 36 and a product of 1428. Find the three terms.
 
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alright so you have two pieces of information here.

you have the fact that a1 * a2 * a3 = 1428

and if d is the difference between a1-a2 and a2-a3,

then you have; a1* (a1 + d) * (a1 + 2d) = 1428

so that's the first piece of information. Do you think you could put the second piece of information in terms of a1 and d, and then you would have 2 variables and 2 equations.
 
Brilliant. Thanks very much for your help.
 
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