the answer is 4,5,6,7

please show me the method.

let four consecutive integers are [a-3d],[a-d],[a+d],[a+3d]

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the answer is 4,5,6,7

please show me the method.

let four consecutive integers are [a-3d],[a-d],[a+d],[a+3d]

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I may be wrong. its been 30 years since last using algebraic expressions [x(x+1)]*[(x+2)*(x+3)]=840 or [x(x-1)]*[(x-2)*(x-3)]=840 multiplication of Binomial expressions

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The four consecutive numbers are (a-3d),a-d,(a+d),a-3d.

As here difference is 2d.but numbers are consecutive, so difference must be 1.

Therefore, 2d =1 i.e. d=0.5

(a-1.5)(a-0.5)(a+0.5)(a+1.5)=840

a^4 - 2.5a^2 - 0.5605=840

Thus gives us a=5.5

Hence, a-3d =5.5 - 1.5 =4

a-d =5.5-0.5 =5

a+d =5.5+0.5 =6

And a+3d =5.5+1.5 =7

Therefore the four numbers are 4,5,6,7

As here difference is 2d.but numbers are consecutive, so difference must be 1.

Therefore, 2d =1 i.e. d=0.5

(a-1.5)(a-0.5)(a+0.5)(a+1.5)=840

a^4 - 2.5a^2 - 0.5605=840

Thus gives us a=5.5

Hence, a-3d =5.5 - 1.5 =4

a-d =5.5-0.5 =5

a+d =5.5+0.5 =6

And a+3d =5.5+1.5 =7

Therefore the four numbers are 4,5,6,7

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