Question

The terms $x_5 = -4$, $x_1, x_2, \dots, x_{100}$ are in an arithmetic progression (AP). It is also given that $2x_6 + 2x_9 = x_{11} + x_{13}$. Find $x_{100}$.

• A. -194
• B. 206
• C. 204
• D. -196

Answer 1

Answer: (A)

-194

Answer 2

Step 1: General formula for terms in an AP
The general term of an AP is:
$x_n = x_1 + (n-1)d,$
where:
$x_1$ is the first term, and
$d$ is the common difference.
Substituting $n = 5$, we get:
$x_5 = x_1 + 4d = -4. \quad \text{(Equation 1)}$
Step 2: Rewrite the given condition in terms of $x_1$ and $d$
The terms $x_6, x_9, x_{11}, x_{13}$ in the AP are:
$x_6 = x_1 + 5d, \quad x_9 = x_1 + 8d, \quad x_{11} = x_1 + 10d, \quad x_{13} = x_1 + 12d.$
The given condition is:
$2x_6 + 2x_9 = x_{11} + x_{13}.$
Substitute the expressions for the terms:
$2(x_1 + 5d) + 2(x_1 + 8d) = (x_1 + 10d) + (x_1 + 12d).$
Simplify:
$2x_1 + 10d + 2x_1 + 16d = x_1 + 10d + x_1 + 12d.$
Combine terms:
$4x_1 + 26d = 2x_1 + 22d.$
Simplify further:
$2x_1 + 4d = 0. \quad \text{(Equation 2)}$
Step 3: Solve for $x_1$ and $d$
From Equation 2:
$x_1 = -2d.$
Substitute $x_1 = -2d$ into Equation 1:
$-2d + 4d = -4.$
Simplify:
$2d = -4 \implies d = -2.$
Now substitute $d = -2$ into $x_1 = -2d$:
$x_1 = -2(-2) = 4.$
Step 4: Find $x_{100}$
The general term of the AP is:
$x_n = x_1 + (n-1)d.$
Substitute $n = 100$, $x_1 = 4$, and $d = -2$:
$x_{100} = 4 + (100-1)(-2).$
Simplify:
$x_{100} = 4 + 99(-2),$
$x_{100} = 4 - 198,$
$x_{100} = -194.$
Final Answer
The 100th term of the AP is:
$\boxed{-194}.$