MCQMediumJEE 2024Sum of Series

JEE Mathematics 2024 Question with Solution

Let ABC be an equilateral triangle. A new triangle is formed by joining the midpoints of all sides of the triangle ABC, and the same process is repeated infinitely many times. If PP is the sum of the perimeters and QQ is the sum of areas of all the triangles formed in this process, then:

  • A

    P2=363QP^2 = 36\sqrt{3}Q

  • B

    P2=63QP^2 = 6\sqrt{3}Q

  • C

    P=363Q2P = 36\sqrt{3}Q^2

  • D

    P2=723QP^2 = 72\sqrt{3}Q

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: An equilateral triangle of side aa is repeatedly replaced by the triangle formed by joining the midpoints of its sides.

Find: The relation between PP and QQ.

An equilateral triangle with an inner medial triangle formed by joining midpoints; outer side labels are a and the inner top segment is labeled a/2.

The side lengths of successive triangles are:

a, a2, a4, a,\ \frac{a}{2},\ \frac{a}{4},\ \dots

So their perimeters form a geometric series:

P=3a+3a2+3a4+=3a(1+12+14+)P = 3a + 3\cdot\frac{a}{2} + 3\cdot\frac{a}{4} + \cdots = 3a\left(1 + \frac{1}{2} + \frac{1}{4} + \cdots\right)

Using the sum of the infinite geometric series,

P=3a1112=6aP = 3a\cdot\frac{1}{1-\frac{1}{2}} = 6a

The area of an equilateral triangle of side aa is:

34a2\frac{\sqrt{3}}{4}a^2

Hence the areas of successive triangles are:

34a2, 316a2, 364a2, \frac{\sqrt{3}}{4}a^2,\ \frac{\sqrt{3}}{16}a^2,\ \frac{\sqrt{3}}{64}a^2,\ \dots

Therefore,

Q=34a2(1+14+116+)Q = \frac{\sqrt{3}}{4}a^2\left(1 + \frac{1}{4} + \frac{1}{16} + \cdots\right)

Again using the sum of the infinite geometric series,

Q=34a21114=33a2Q = \frac{\sqrt{3}}{4}a^2\cdot\frac{1}{1-\frac{1}{4}} = \frac{\sqrt{3}}{3}a^2

Now from

P=6aP = 6a

we get

a=P6a = \frac{P}{6}

Substituting into QQ,

Q=33(P6)2=3108P2Q = \frac{\sqrt{3}}{3}\left(\frac{P}{6}\right)^2 = \frac{\sqrt{3}}{108}P^2

So,

P2=363QP^2 = 36\sqrt{3}Q

Therefore, the correct option is A.

Geometric Series Approach

Given: Each new triangle is obtained by joining the midpoints of the sides of the previous equilateral triangle.

Find: A direct relation between P2P^2 and QQ.

By the midpoint theorem, each new triangle has side half of the previous one. Therefore the perimeters are in the ratio 1:12:14:1: \frac{1}{2}: \frac{1}{4}: \dots and the areas are in the ratio 1:14:116:1: \frac{1}{4}: \frac{1}{16}: \dots.

Perimeter sum:

P=3a(1+12+14+)=3a2=6aP = 3a\left(1 + \frac{1}{2} + \frac{1}{4} + \cdots\right) = 3a\cdot 2 = 6a

Hence,

P2=36a2P^2 = 36a^2

Area sum:

Q=34a2(1+14+116+)Q = \frac{\sqrt{3}}{4}a^2\left(1 + \frac{1}{4} + \frac{1}{16} + \cdots\right)

Since

1+14+116+=1114=431 + \frac{1}{4} + \frac{1}{16} + \cdots = \frac{1}{1-\frac{1}{4}} = \frac{4}{3}

we get

Q=34a243=33a2Q = \frac{\sqrt{3}}{4}a^2\cdot\frac{4}{3} = \frac{\sqrt{3}}{3}a^2

Therefore,

P2Q=36a233a2=363\frac{P^2}{Q} = \frac{36a^2}{\frac{\sqrt{3}}{3}a^2} = 36\sqrt{3}

Hence,

P2=363QP^2 = 36\sqrt{3}Q

This matches option A. Note: the line in the source solution stating "Answer: (1)  P=363Q(1)\; P = 36\sqrt{3}Q" is a typo; the derived result and the option list both confirm P2=363QP^2 = 36\sqrt{3}Q.

Common mistakes

  • Students often assume the area also scales by 12\frac{1}{2} because the side scales by 12\frac{1}{2}. This is wrong because area depends on the square of the side. Use area ratio (12)2=14\left(\frac{1}{2}\right)^2 = \frac{1}{4} instead.

  • A common mistake is to omit the original triangle from the sums PP and QQ. The process includes the first triangle as the first term of both geometric series, so begin with 3a3a for perimeter and 34a2\frac{\sqrt{3}}{4}a^2 for area.

  • Some students use the wrong infinite series formula or interchange the common ratios for perimeter and area. The perimeter series has ratio 12\frac{1}{2}, while the area series has ratio 14\frac{1}{4}. Keep these two series separate before relating PP and QQ.

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