If a, b, c, d are distinct integers in A. P. Such that `d=a^2+b^2+c^2`, then a + b + c + d is

Explanation

Here, sum of numbers `i.e.,a+b+c+d` is not given.
Let `b=a+D,c=a+2D, d=a+3D,AAD in N`
According to hypothesis,
`a+3D=a^(2)+(a+D^(2))+(a+2D)^(2)`
` implies 5D^(2)+3(2a-1)D+3a^(2)-a=0" " ".....(i)"`
` therefore D=(-3(2a-1)pm sqrt(9(2a-1)^(2)-20(3a^(2)-a)))/(10)`
` =(-3(2a-1)pm sqrt(24a^(2)-16a+9))/(10)`
Now, ` -24a^(2)-16a+9 ge 0`
` implies 24a^(2)-16a-9 le 0` ` implies -(1)/(3)-(sqrt70)/(3) le-(1)/(3)+(sqrt70)/(12)`
` implies a=-1,0 " " [therefore a in I]`
When a=0 from Eq.(i), `D=0,(3)/(5)("not possible"therefore D in N) " and for " a=-1`
From Eq. (i), `D=1,(4)/(5)`
`therefore D=1`
`therefore a=-1, b=0, c=1, d=2 " " [therefore D in N]`
Then, `a+b+c+d=1-+0+1+2=2`