Four different integers form an increasing A.P. One of these numbers is equal to the sum of the squares of the other three numbers. ThenThe product of all numbers isThe sum of all the four numbers isThe common difference of the four numbers is

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Let the four integers be a $$-$$ d, a, a $$+$$ d, and a $$+$$ $$2d$$, where aand d are integers and d > $$0$$. Now,$$a+2d=(a$$−$$d)2+a2+(a+d)22d2$$−$$2d+3a2$$−$$a=0$$ $$----1d=121$$±$$1+2a$$−$$6a2-----2Since$$ d is a positive integer, $$so1+2a$$−$$6a2$$>$$0or$$ $$6a2$$−$$2a$$−$$1$$<$$0or$$ $$1$$−$$76$$ $$0$$, d $$=$$ $$1Hence$$, the four numbers are,$$-1$$, $$0$$, $$1$$,$$2$$.Let the four integers be a $$-$$ d, a, a $$+$$ d, and a $$+$$ $$2d$$, where aand d are integers and d > $$0$$. Now,$$a+2d=(a$$−$$d)2+a2+(a+d)22d2$$−$$2d+3a2$$−$$a=0$$ $$----1d=121$$±$$1+2a$$−$$6a2-----2Since$$ d is a positive integer, $$so1+2a$$−$$6a2$$>$$0or$$ $$6a2$$−$$2a$$−$$1$$<$$0or$$ $$1$$−$$76$$ $$0$$, d $$=$$ $$1Hence$$, the four numbers are,$$-1$$, $$0$$, $$1$$,$$2$$.Let the four integers be a $$-$$ d, a, a $$+$$ d, and a $$+$$ $$2d$$, where aand d are integers and d > $$0$$. Now,$$a+2d=(a$$−$$d)2+a2+(a+d)22d2$$−$$2d+3a2$$−$$a=0$$ $$----1d=121$$±$$1+2a$$−$$6a2-----2Since$$ d is a positive integer, $$so1+2a$$−$$6a2$$>$$0or$$ $$6a2$$−$$2a$$−$$1$$<$$0or$$ $$1$$−$$76$$ $$0$$, d $$=$$ $$1Hence$$, the four numbers are,$$-1$$, $$0$$, $$1$$,$$2$$.

Four different integers form an increasing A.P. One of these numbers is equal to the sum of the squares of the other three numbers. ThenThe product of all numbers isThe sum of all the four numbers isThe common difference of the four numbers is

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