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The sum of all the four numbers is |
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Answer» 3 `a+2d=(a-d)^(2)+a^(2)+(a+d)^(2)` or `2d^(2)-2d+3a^(2)-a=0`(1) `therefored=1/2[1pmsqrt(1+2a-6a^(2))]`(2) Since d is a positive INTEGER, so `1+2a-6a^(2)gt0` or `6a^(2)-2a-1lt0` or `(1-sqrt7)/6ltalt(1+sqrt7)/6` or a=0 (`because` a is an integer) Hence, from (2), d=1 or 0 But since `dgt0`, `therefore` d=1 Hence, the four NUMBERS are -1,0,1,2. |
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