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Official CAT 2018 Questions; Section: QA

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.



sum infinite terms of a Geometric Progression

Martes 6 de agosto de 2019, por hurtado claudio

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

1) 248 √3
2) 192 √3
3) 188 √3
4) 164 √3


If we start from an equilateral triangle of side "a", and join the midpoints of its sides, an equilateral triangle of side "a / 2" is formed and so on ...
In an equilateral triangle of side "a", the area is (a ^ 2 (squr 3)) / 4.
In an equilateral triangle of side "a / 2", the area is (a ^ 2 (sqr3)) / 16.
In an equilateral triangle of side "a / 4", the area is
(a ^ 2 (sqr3)) / 64
and so on...

We have to repeat in each area a ^ 2 (sqr3)

Factoring we have:

(a ^ 2 (sqr3)) (1/4 + 1 / (4 ^ 2) + 1 / (4 ^ 3) + 1 / (4 ^ 4) + ... + 1 / (4 ^ n) +. ....)

Where (1/4 + 1 / (4 ^ 2) + 1 / (4 ^ 3) + 1 / (4 ^ 4) + ... + 1 / (4 ^ n) + .....), is the sum infinite terms of a Geometric Progression, of ratio 1/4.

IF WE ADD THE INFINITE TERMS of a GP whose reason, "r" is between zero and one, you get:
(The first adding) / (1-r).

Then
(1/4 + 1 / (4 ^ 2) + 1 / (4 ^ 3) + 1 / (4 ^ 4) + ... + 1 / (4 ^ n) + .....) = (1 / 4) / (1- (1/4)) = (1/4) / (3/4) = (1/3)

Finally we have
The sum requested, for any initial triangle from side a, is (a ^ 2 (sqr3)) (1/3).

For our situation
(24 ^ 2 (sqr3) (1/3) = 192 (sqr3)

Answer 2)

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