An algebra problem by Rajsuryan Singh

It happens again. I solve the problem but it shows that I did not.

Anyway, the solution is 2 by observation and then I calculated it as follows.

Since $d=a^2+b^2+c^2$ , this implies that $d$ is the fourth term $a_4$ .

Let the first, second and third terms be $a$ , $b$ and $c$ , then $b-a=c-b=x$ , where x is the difference between two successive terms.

Then $a=b-x$ , $c=b+x$ and $d=b+2x$ .

Therefore,

$d=b+2x=(b-x)^2+b^2+(b+x)^2$

$\Rightarrow b+2x = b^2-2bx+x^2+b^2+b^2+2bx+x^2$

$\quad \quad b+2x = 3b^2+2x^2$

$\quad \quad 2x^2-2x+(3b^2-b)=0$

$\Rightarrow x=\dfrac{2 \pm \sqrt{4-8(3b^2-b)} }{4}$

$\quad \quad x=\dfrac{2 \pm 2\sqrt{1+2b-24b^2)} }{4}$

Note that $x = 2n$ , when $n$ is a positive integer.

For $x=2n$ , implies that $\sqrt{1+2b-24b^2}$ is an integer $\Rightarrow b=0$ and $x = 1$ .

Therefore,

$a=-1\quad b = 0\quad c=1\quad d=2$

$\Rightarrow a+b+c+d=-1+0+1+2=\boxed{2}$

Sorry guys, i don't have any appropriate solution from this problem. But maybe it can help this:

1st, Take a note that the series is in arithmetic progression so that means, if $a$ is the first term then the second term must be $a+d$ , 3rd is $a+2d)$ and and so on...

2nd, From the equation $d={ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$ , the series must be a smaller integer because if the integer is larger, then it not might suitable to solve the equation.

3rd, By solving this, i assume that the middle $b$ is zero. And it works. Possible solution is $2={ (-1) }^{ 2 }+{ 0 }^{ 2 }+{ 1 }^{ 2 }$

Answer : $-1+0+1+2=\boxed{2}$

SOMETIMES , we can solve the problem through our analysis by doing Assumption .

try it by hit and error method i considered the AP -1,0,+1,+2 and it satisfied the condition. so sum is 2

$a = x - e$

$b = x$

$c = x + e$

$d = x + 2e$

We have equation to find the relationship between x and e:

$x+2e = (x-e)^2 + x^2 + (x+e)^2$

$x + 2e = 3x^2 + 2e^2$

But since x, e are integers :

$x <= 3x^2 ("=" <=> x = 0)$

$2e<= 2e^2 ("=" <=> (e = 0) or (e = 1))$

but e must = 1 to make integers distinct.

So we only have (x,e) = (0,1) sum = 4x + 2e = 2.

first solve give equation d=a2 +b2+c2 a2+b2+c2-d=0 we know that 1 a2+1 b2+1 c2+(-1) d=0 we found coefficent of this equation is a=1,b=1,c=1,d=-1 we put this value in finding equation 1+1+1-1 =2 this is the answer