Question

$x<-1$ and $x,|x+1|,|x-1|$ are in A.P. The sum of the first twenty terms of the A.P. is

• A. $150$
• B. $280$
• C. $350$
• D. $180$

Answer 1

Answer: (C)

$350$

$x<-1$

$x+1<-1+1$

$x+1<0$

So, $|x+1|=-(x+1)$

$|x+1|=-x-1$ .................... (1)

$x<-1$

$x-1<-1-1$

$x-1<-2$

So, $|x-1|=-(x-1)$

$|x-1|=-x+1$ ...................... (2)

Now, $x,-x-1,-x+1$ are in A.P.

So,

$2(-x-1)=x+(-x+1)$

$2(-x-1)=1$

$-x-1=\frac{1}{2}$

First term, $x=-\frac{1}{2}-1=-\frac{3}{2}$

Second term, $|x+1|=|-\frac{3}{2}+1|=\frac{1}{2}$

Common difference, $d=\frac{1}{2}-(-\frac{3}{2})=2$

Sum of $n$ terms, $S_n=\frac{n}{2}\{2a+(n-1\left)d\right\}$

$S_{20}=10\{2a+19d\}$

$S_{20}=10\{2(-\frac{3}{2})+19\times 2\}$

$S_{20}=10(-3+38)$

$S_{20}=10\left(35\right)=350$