Question

11. The sum of three consecutive terms that are in A.P. is 27 and their product is 288.
Find the three terms.
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Answer 1

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Answer 2

Answer:

2,9 and 16

Step-by-step explanation:

let the three terms be a-d,a,a+d

according to first condition,

[a-d]+a+[a+d]=27

3a=27

a=9

according to second condition,

[a-d][a][a+d]=288

[a-d][a*a+ad]=288

a[a*a+ad]-d[a*a+ad]=288

a3+a2d-a2d-ad2=288

∴a3-ad2=288

∴729-ad2=288........[a=9]

-ad2=288-729

-ad2=-441

ad2=441

∴d2=441÷9[a=9]

d2=49

∴d=7

∴first term=a-d=9-7=2

∴second term=a=9

∴third term=a+d=9+7=16