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Question:

Given an equilateral triangle T₁ with side 24 cm, a second triangle T₂ is formed by joining the midpoints of the sides of T₁. Then a third triangle T₃ is formed by joining the midpoints of the sides of T₂. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T₁, T₂, T₃,... will be

Explanation:

Consider ∆ABC (i.e. T₁) and ∆DEF (i.e., T₂).

D and E are midpoints of AB and AC respectively. Therefore, BC = 2 × DE

Side of T₂ = 1/2 × Side of T₁

Area of T₁ = A(T₁) = √3/4 × 24²

Similarly, A(T₂) = √3/4 × 12²

A(T₃) = √3/4 × 6² ... and so on

Sum of areas of infinitely many T_i’s

= √3/4 × 24² + √3/4 × 12² + √3/4 × 6² + ...

= √3/4 (24² + 12² + 6² + ...)

Here, (24² + 12² + 6² + ...) is an infinite series with r = 1/4

Hence, = √3/4 (24² + 12² + 6² + ...) = $\frac{\sqrt{3}}{4}\left(\frac{{24}^2}{\left(1-{\displaystyle \frac{1}{4}}\right)}\right)$ = 192√3

Hence, option (d).