JEE Main 2024 (Online) 6th April Evening ShiftQuestion 1
Let $ABC$ be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle $ABC$ and the same process is repeated infinitely many times. If $P$ is the sum of perimeters and $Q$ is the sum of areas of all the triangles formed in this process, then :
Sequence and Series · Geometric Progression
Solution
Concept. Joining the midpoints halves every side. So perimeters form a G.P. with ratio $\tfrac12$ and areas a G.P. with ratio $\tfrac14$. Formulas used. (1) Area of an equilateral triangle of side $a$: $\frac{\sqrt3}{4}a^2$(2) Midpoint theorem: each side of the new triangle is half a side of the old one(3) Sum of an infinite G.P.: $\frac{a}{1-r}$ for $|r|<1$Step 1 - The figure. Let the side of $ABC$ be $a$.The sides are $a,\ \dfrac a2,\ \dfrac a4,\dots$ Step 2 - Sum of perimeters. $$P=3a\left(1+\frac12+\frac14+\cdots\right)=\frac{3a}{1-\frac12}=6a .$$ Step 3 - Sum of areas. $$Q=\frac{\sqrt3}{4}a^2\left(1+\frac14+\frac1{16}+\cdots\right)=\frac{\sqrt3}{4}a^2\cdot\frac43=\frac{a^2}{\sqrt3}.$$ Step 4 - Relation. $a^2=\sqrt3\,Q$, so $$P^2=36a^2=36\sqrt3\,Q .$$ Answer: (C)