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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Sequences and Series

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Sequences and Series

Let A B C be an equilateral triangle. A new triangle is formed by...

Let ABCA B C be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABCA B C and the same process is repeated infinitely many times. If P\mathrm{P} is the sum of perimeters and QQ is be the sum of areas of all the triangles formed in this process, then :

Options

  1. AP2=63Q\mathrm{P}^2=6 \sqrt{3} \mathrm{Q}
  2. BP2=363Q\mathrm{P}^2=36 \sqrt{3} \mathrm{Q}
  3. CP=363Q2\mathrm{P}=36 \sqrt{3} \mathrm{Q}^2
  4. DP2=723Q\mathrm{P}^2=72 \sqrt{3} \mathrm{Q}

Correct answer

B. P2=363Q\mathrm{P}^2=36 \sqrt{3} \mathrm{Q}

Step-by-step solution

Area of first Δ=3a24\Delta=\frac{\sqrt{3} \mathrm{a}^2}{4} Area of second Δ=3a24a24=3a216\Delta=\frac{\sqrt{3} \mathrm{a}^2}{4} \frac{\mathrm{a}^2}{4}=\frac{\sqrt{3} \mathrm{a}^2}{16} Area of third Δ=3a264\Delta=\frac{\sqrt{3} \mathrm{a}^2}{64} sum of area =3a24=\frac{\sqrt{3} \mathrm{a}^2}{4}\left(1+\frac{1}{4}+\frac{1}{16}\ldots\right) Q=3a24134=a23 perimeter of 1st Δ=3a perimeter of 2nd Δ=3a2 perimeter of 3rd Δ=3a4P=3a(1+12+14+…)P=3a⋅2=6aa=P6Q=13⋅P236P2=363Q\begin{aligned} & \mathrm{Q}=\frac{\sqrt{3} \mathrm{a}^2}{4} \frac{1}{\frac{3}{4}}=\frac{\mathrm{a}^2}{\sqrt{3}} \\ & \text{ perimeter of } 1^{\text{st }} \Delta=3 \mathrm{a} \\ & \text{ perimeter of } 2^{\text{nd }} \Delta=\frac{3 \mathrm{a}}{2} \\ & \text{ perimeter of } 3^{\text{rd }} \Delta=\frac{3 \mathrm{a}}{4} \\ & \mathrm{P}=3 \mathrm{a}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right) \\ & \mathrm{P}=3 \mathrm{a} \cdot 2=6 \mathrm{a} \\ & \mathrm{a}=\frac{\mathrm{P}}{6} \\ & \mathrm{Q}=\frac{1}{\sqrt{3}} \cdot \frac{\mathrm{P}^2}{36} \\ & \mathrm{P}^2=36 \sqrt{3} \mathrm{Q}\end{aligned}
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