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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

$0$

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B

$1$

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C

$2$

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D

None of these

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Solution

## 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 $A.P$.$⇒$ $A+2D={\left(A-D\right)}^{2}+{A}^{2}+{\left(A+D\right)}^{2}$$⇒$ $A+2D={A}^{2}-2AD+{D}^{2}+{A}^{2}+{A}^{2}+2AD+{D}^{2}$$⇒$ $A+2D=3{A}^{2}+2{D}^{2}$$⇒$ $3{A}^{2}+2{D}^{2}-A-2D=0$$⇒$$2{D}^{2}-2D+\left(3{A}^{2}-A\right)=0$$⇒$ $D=\frac{\left[2±\sqrt{\left(4-8\left(3{A}^{2}-A\right)\right)}\right]}{4}$ $=\frac{\left[1±\sqrt{\left(1-2\left(3{A}^{2}-A\right)}\right]}{2}\phantom{\rule{0ex}{0ex}}=\frac{\left[1±\sqrt{\left(1-6{A}^{2}+2A\right)}\right]}{2}\dots .\left(i\right)$Since $D$ should be an integer, Step 2 : let $\mathbf{1}\mathbf{-}\mathbf{6}{\mathbit{A}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbit{A}\mathbf{}\mathbf{=}\mathbf{}{\mathbit{m}}^{\mathbf{2}}$ multiply by $6$$6\left(1-6{A}^{2}+2A\right)=6{m}^{2}$$⇒$ ${\left(6A-1\right)}^{2}=7-6{m}^{2}$LHS is positive, so $7-6{m}^{2}>0$$⇒$$m=0or±1$$a+b+c+d=A-D+A+A+D+A+2D$ $=4A+2D\dots \left(ii\right)$$⇒$ $D=\left(1±m\right)/2$When $m=0$$\begin{array}{rcl}D& =& \frac{\left(1+0\right)}{2}\\ & =& \frac{1}{2}\end{array}$When $m=1$$\begin{array}{rcl}D& =& \frac{\left(1+1\right)}{2}\\ & =& 1\end{array}$When $m=-1$$\begin{array}{rcl}D& =& \left(1-1\right)/2\\ & =& 0\end{array}$$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=\left[1±\sqrt{\left(1-6{A}^{2}+2A\right)}\right]/2$$⇒$ $2=1±\sqrt{\left(1-6{A}^{2}+2A\right)}$$⇒$ $1=1-6{A}^{2}+2A\right)$$⇒$ $6{A}^{2}–2A=0$$⇒$$2A\left(3A-1\right)=0$$⇒$$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.

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