TheCATExam
Sign in

CAT 2024 Slot 1 — QA Question 17

Arithmetic ProgressionEasy

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

Answer & solution

  • -194

  • B

    206

  • C

    204

  • D

    -196

Solution

Medium

Write every term as x1+(n−1)dx_1+(n-1)d. The condition 2x6+2x9=x11+x132x_6+2x_9=x_{11}+x_{13} is linear in dd and forces a specific dd; combine with x5=−4x_5=-4 to get x1x_1, then evaluate x100x_{100}.

1

Use the index condition. Replace each term by x1+(n−1)dx_1+(n-1)d.

2x6+2x9=x11+x13⇒ 2(x1+5d)+2(x1+8d)=(x1+10d)+(x1+12d)⇒ 4x1+26d=2x1+22d(collect)⇒ 2x1=−4d ⇒ x1=−2d\begin{aligned} &2x_6+2x_9=x_{11}+x_{13}\\ &\Rightarrow\ 2(x_1+5d)+2(x_1+8d)=(x_1+10d)+(x_1+12d)\\ &\Rightarrow\ 4x_1+26d=2x_1+22d \quad\text{(collect)}\\ &\Rightarrow\ 2x_1=-4d\ \Rightarrow\ x_1=-2d \end{aligned}
2

Use x5=−4x_5=-4. Substitute x1=−2dx_1=-2d from step 1.

x5=x1+4d=−4⇒ −2d+4d=−4(substitute step 1)⇒ 2d=−4 ⇒ d=−2, x1=4\begin{aligned} &x_5=x_1+4d=-4\\ &\Rightarrow\ -2d+4d=-4 \quad\text{(substitute step 1)}\\ &\Rightarrow\ 2d=-4\ \Rightarrow\ d=-2,\ x_1=4 \end{aligned}
3

Evaluate x100x_{100} with x1=4, d=−2x_1=4,\ d=-2.

x100=x1+99d=4+99(−2)⇒ x100=4−198(from step 2)⇒ x100=−194\begin{aligned} &x_{100}=x_1+99d=4+99(-2)\\ &\Rightarrow\ x_{100}=4-198 \quad\text{(from step 2)}\\ &\Rightarrow\ x_{100}=-194 \end{aligned}
x100=−194x_{100}=-194
CAT 2024 Slot 1 QA Q17: Suppose x 1 , x 2 , x 3 , ..., x 100 are in arithmetic progression such that x 5 = -4 and 2x 6 + 2x 9 = x 11 + — Solution | TheCATExam