Suppose x₁, x₂, x₃,…, x100 are in arithmetic progression such that x₅ = –4 and 2x₆ + 2x₉ = x₁₁ + x₁₃, Then, x₁₀₀ equals

The correct option is –194.

Step-by-step Explanation:

Let the first term of the arithmetic progression (A.P.) be $a$ and the common difference be $d$.
The general formula for the $n$-th term of an A.P., denoted by $x_n$, is given by:

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

According to the problem, we are given the following two conditions:

Condition 1:
The fifth term is $-4$:

$$x_5=a+4d=-4$$

Let us label this as Equation (1).

Condition 2:
We are given the relation:

$$2x_6+2x_9=x_{11}+x_{13}$$

Expressing each term in this relation using the general formula:

$$x_6=a+5d$$

$$x_9=a+8d$$

$$x_{11}=a+10d$$

$$x_{13}=a+12d$$

Substituting these expressions back into the given relation:

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

Now, simplify both sides of the equation:
Left-hand side (LHS):

$$2a+10d+2a+16d=4a+26d$$

Right-hand side (RHS):

$$a+10d+a+12d=2a+22d$$

Equating LHS and RHS:

$$4a+26d=2a+22d$$

Grouping the terms together:

$$4a-2a+26d-22d=0$$

$$2a+4d=0$$

Dividing by 2 gives:

$$a+2d=0\Rightarrow a=-2d$$

Let us label this as Equation (2).

Solving for $a$ and $d$:
Substitute Equation (2) into Equation (1):

$$(-2d)+4d=-4$$

$$2d=-4$$

$$d=-2$$

Now find the value of $a$ using Equation (2):

$$a=-2(-2)=4$$

Finding the 100th term ($x_{100}$):
Using the term formula for $n=100$:

$$x_{100}=a+99d$$

Substituting the values of $a$ and $d$:

$$x_{100}=4+99(-2)$$

$$x_{100}=4-198$$

$$x_{100}=-194$$

Therefore, the value of $x_{100}$ is $-194$.