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

Accepted Answer

B. Let first term be $a$ and common difference be $d$. $x_5 = a + 4d = -4$. Given $2x_6 + 2x_9 = x_{11} + x_{13}$: $$2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d)$$ $$4a + 26d = 2a + 22d \implies 2a + 4d = 0 \implies a = -2d$$ Substitute $a = -2d$ into $a + 4d = -4$: $$-2d + 4d = -4 \implies 2d = -4 \implies d = -2$$ $$a = 4$$ Then $x_{100} = a + 99d = 4 + 99(-2) = -194$.

Arithmetic Progression Term

Explanation

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