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 ?
- 206
- -196
- 204
- -194
The Correct Option is D
Solution and Explanation
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 $X_{100}$:
$X_{100} = a + 99d = -2(-2) + 99(-2) = 4 - 198 = -194$
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