CAT 2024 Slot 1 QA Question 21: Suppose $x_1, x_2, x_3, \ldots, x_{100}$ are in arithmetic progression such that $x_5 = -4$ and $2x_6 + 2x_9 = x_{11} + x_{13}$. Then, $x_{100}$ equals

Question

Suppose $x_1, x_2, x_3, \ldots, x_{100}$ are in arithmetic progression such that $x_5 = -4$ and $2x_6 + 2x_9 = x_{11} + x_{13}$. Then, $x_{100}$ equals

Accepted Answer

Let $x_1 = a$ and the common difference be $d$.
$x_5 = a + 4d = -4$
Use the second condition.
$2x_6 + 2x_9 = x_{11} + x_{13}$
$2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d)$
$\Rightarrow 4a + 26d = 2a + 22d$
$\Rightarrow a + 2d = 0$
$a + 4d = -4$ and $a + 2d = 0$
$\Rightarrow 2d = -4$
$\Rightarrow d

Suppose x1,x2,x3,…,x100x_1, x_2, x_3, \ldots, x_{100}x1​,x2​,x3​,…,x100​ are in arithmetic progression such that x5=−4x_5 = -4x5​=−4 and 2x6+2x9=x11+x132x_6 + 2x_9 = x_{11} + x_{13}2x6​+2x9​=x11​+x13​. Then, x100x_{100}x100​ equals

Suppose $x_1, x_2, x_3, \ldots, x_{100}$ are in arithmetic progression such that $x_5 = -4$ and $2x_6 + 2x_9 = x_{11} + x_{13}$ . Then, $x_{100}$ equals