Question

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

• A. 206
• B. -196
• C. 204
• D. -194

Answer 1

Answer: (D)

-194

Answer 2

Let the first term of the arithmetic progression be $a$ and the common difference be $d$.
Thus:

$X_n = a + (n - 1)d$

From the given conditions:
- $X_5 = a + 4d = -4$
- $2X_6 + 2X_9 = X_{11} + X_{13}$

Using the formula for terms:
- $X_6 = a + 5d$ - $X_9 = a + 8d$ - $X_{11} = a + 10d$ - $X_{13} = a + 12d$

Substitute into the equation:

$2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d)$

Simplifying:

$2a + 10d + 2a + 16d = 2a + 22d$
$4a + 26d = 2a + 22d$
$2a + 4d = 0 \implies a = -2d$

Substitute $a = -2d$ into $X_5 = -4$:

$-2d + 4d = -4 \implies 2d = -4 \implies d = -2$

Now, find $X_{100}$:

$X_{100} = a + 99d = -2(-2) + 99(-2) = 4 - 198 = -194$