" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/057af284a2b6.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Sequences and Series Question 64 English Explanation">

$$\triangle A B C$$ is an equilateral triangle having side $$=a$$ unit

Now, perimeter $$=$$ sum of all sides $$=3 a$$

Area $$=\frac{\sqrt{3}}{4} a^2$$

Now, $$\ln \triangle D E F, D E=\frac{a}{2}=E F=D F$$

$$\begin{aligned} & \text { Perimeter }=3 \times \frac{a}{2}=\frac{3 a}{2} \\ & \text { Area }=\frac{\sqrt{3}}{4} \times\left(\frac{a}{2}\right)^2=\frac{\sqrt{3} a^2}{16} \end{aligned}$$

Now, $$P=3 a+\frac{3 a}{2}+\frac{3 a}{4}+\cdots$$

$$Q=\frac{\sqrt{3}}{4} a^2+\frac{\sqrt{3}}{16} a^2+\frac{\sqrt{3}}{64} a^2+\cdots$$

$$P=\frac{3 a}{1-\frac{1}{2}}=3 a \times 2 \Rightarrow P=6 a \quad \text{.... (i)}$$

$$Q=\frac{\frac{\sqrt{3}}{4} a^2}{1-\frac{1}{4}}=\frac{4}{3} \times \frac{\sqrt{3}}{4} a^2 \quad Q=\frac{\sqrt{3}}{3} a^2 \quad \text{.... (ii)}$$

From equation (i) & (ii)

$$\begin{aligned} & P=6 a \\ & Q=\frac{\sqrt{3}}{3} a^2 \\ & Q=\frac{\sqrt{3}}{3} \times \frac{P^2}{36} \\ & P^2=36 \sqrt{3} Q \end{aligned}$$

19:T6a3,

JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Sequences and Series Question 64 Hindi Explanation

$$\triangle A B C$$ एक समबाहु त्रिभुज है जिसकी भुजा $$=a$$ इकाई है

अब, परि

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