Question

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

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Step 1 : Let $A-D,A,A+D,A+2D$be the distinct terms in $A.P$.

$\Rightarrow$ $A+2D={(A-D)}^2+A^2+{(A+D)}^2$

$\Rightarrow$ $A+2D=A^2-2AD+D^2+A^2+A^2+2AD+D^2$

$\Rightarrow$ $A+2D=3A^2+2D^2$

$\Rightarrow$ $3A^2+2D^2-A-2D=0$

$\Rightarrow$$2D^2-2D+(3A^2-A)=0$

$\Rightarrow$ $D=\frac{[2\pm \sqrt{(4-8(3A^2-A\left)\right)}]}{4}$

$$\begin{gathered} =\frac{[1\pm \sqrt{(1-2(3A^2-A)}]}{2} \\ =\frac{[1\pm \sqrt{(1-6A^2+2A)}]}{2}\dots .\left(i\right) \end{gathered}$$

Since $D$ should be an integer,

Step 2 : let $\mathbf{1}\mathbf{-}\mathbf{6}{\mathit{A}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}{\mathit{m}}^{\mathbf{2}}$ multiply by $6$

$6(1-6A^2+2A)=6m^2$

$\Rightarrow$ ${(6A-1)}^2=7-6m^2$

LHS is positive, so $7-6m^2>0$

$\Rightarrow$$m=0or\pm 1$

$a+b+c+d=A-D+A+A+D+A+2D$

$=4A+2D\dots \left(ii\right)$

$\Rightarrow$ $D=(1\pm m)/2$

When $m=0$

$\begin{array}{rcl}D& =& \frac{(1+0)}{2}\\ & =& \frac{1}{2}\end{array}$

When $m=1$

$\begin{array}{rcl}D& =& \frac{(1+1)}{2}\\ & =& 1\end{array}$

When $m=-1$

$\begin{array}{rcl}D& =& (1-1)/2\\ & =& 0\end{array}$

$D$ should be an integer.

So $D=1or0$

Since $a,b,c,d$are distinct integers, $D$ cannot be $0$.

So $D=1$

Step 4. Substitute $D$ in $\left(i\right)$, we get

$1=[1\pm \sqrt{(1-6A^2+2A)}]/2$

$\Rightarrow$ $2=1\pm \sqrt{(1-6A^2+2A)}$

$\Rightarrow$ $1=1-6A^2+2A)$

$\Rightarrow$ $6A^2–2A=0$

$\Rightarrow$$2A(3A-1)=0$

$\Rightarrow$$A=0orA=\frac{1}{3}$

But $A$ cannot be $\frac{1}{3}$ .

Step 5. Put $A=0$ and $D=1$ in $\left(ii\right)$, we get

$\begin{array}{rcl}\therefore a+b+c+d& =& 4A+2D\\ & =& 0+2\\ & =& \mathbf{2}\end{array}$

Hence, Option ‘C’ is Correct.