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

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SUMMARY

The problem involves finding three consecutive terms in an arithmetic progression (A.P.) that sum to 36 and have a product of 1428. The terms can be represented as a1, a1 + d, and a1 + 2d, where d is the common difference. The equations derived from the sum and product are a1 + (a1 + d) + (a1 + 2d) = 36 and a1 * (a1 + d) * (a1 + 2d) = 1428. Solving these equations will yield the specific values of a1 and d, leading to the three terms.

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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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