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 ?

Question

Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to

Answer 1

Let the first term of the arithmetic progression be $a$ and the common difference be $d$. The formula for the $n^{th}$ term is:

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

Given the following conditions:

- $X_5 = a + 4d = -4$

- $2X_6 + 2X_9 = X_{11} + X_{13}$

Using the formula for the terms:

- $X_6 = a + 5d$

- $X_9 = a + 8d$

- $X_{11} = a + 10d$

- $X_{13} = a + 12d$

Substitute these into the second equation:

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

Simplify the equation:

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

Substitute $a = -2d$ into the equation for $X_5$:

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

Now, calculate $X_{100}$:

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

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 ?

The Correct Option is D

Solution and Explanation

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