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The sum of the four numbers which are in arithmetic progression is 20 and product is 625. Find the numbers?

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by lkjh07

let 4 no be a-3d,a-d,a+d,a+3d

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by lkjh07

let 4 no be a-3d,a-d,a+d,a+3d

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Sum of these numbers is

a-2d+a-d+a+d+a+2d=20

we can find a values by cancelling d's so a=5

now product of these values isĀ

(a-2d)*(a-d)*(a+d)*(a+2d)=625

taking a common we get

a[(1-2d)*(1-d)*(1+d)*(1+2d)]=625

since a=5;

after computing we get

1-5d^2+4d^4=125

4d^4-5d^2=124

we can find the d value and on substituting that we can get the four values due to time i cant complete the problem but this is the concept.