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CAT 2024 Slot 1 QA Question & Solution

AlgebraMedium

Question

Suppose $x_{1},x_{2},x_{3},...,x_{100}$ are in arithmetic progression such that $x_{5}=-4$ and $2x_{6}+2x_{9}=x_{11}+x_{13}$, Then,$x_{100}$ equals

Options

-194
-196
204
206

Solution

Using the arithmetic progression formula for the nth term, where
$x_n=a+\left(n-1\right)d$
Substituting the value for n and using that in the equation that is given, 
$2x_{6}+2x_{9}=x_{11}+x_{13}$, Then,$x_{100}$ equals

We get, $2\left(a+5d\right)+2\left(a+8d\right)=a+10d+a+12d$
$4a+26d=2a+22d$
$2a=-4d$
$a=-2d$

We are given, $x_5=-4$
$a+4d=-4$
Substituting the value for a in terms of d, 
$2d=-4$
$d=-2$
$a=4$

$x_{100}=a+99d$
$x_{100}=4-198=-194$