Question

If $x,|x+1|,|x-1|$ are first three terms of an A.P., then the maximum possible sum of the first $20$ terms is

• A. $350$
• B. $180$
• C. $370$
• D. $x,|x+1|,|x-1|$ cannot be first three terms of an A.P. for all $x\in R$

Answer 1

The correct option is A $350$

​​​​​Given $2|x+1|=x+|x-1|$

Case $\left(i\right)$ $x\ge 1$
$2(x+1)=x+x-1\Rightarrow 2=-1$
Which is absurd

Case $\left(ii\right)$ $-1\le x<1$
$2(x+1)=x-(x-1)$
$\Rightarrow x=-\frac{1}{2}$
then $|x+1|=\frac{1}{2}$
$\Rightarrow d=\frac{1}{2}+\frac{1}{2}=1$
Sum of first $20$ terms $=10[-1+19]=180$

Case $\left(iii\right)$ $x<-1$
$-2(x+1)=x-(x-1)$
$\Rightarrow x=-\frac{3}{2}$
then $|x+1|=\frac{1}{2}$
$\Rightarrow d=\frac{1}{2}+\frac{3}{2}=2$
Sum of the first $20$ terms $10[-3+19\cdot 2]=350$