A side of an equilateral triangle is $20$ cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle. ![A sequence of nested equilateral triangles: a large equilateral triangle (side 20 cm), a second equilateral triangle inscribed by joining — Mathematics question](/figures/sequences-and-series/sequence-nested-equilateral-triangles-large-equilateral-triangle-fig-pyq-t6lkt5.svg)

A side of an equilateral triangle is 20 cm long. A second equilateral triangle is inscribed in it by joining t

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A sequence of nested equilateral triangles: a large equilateral triangle (side 20 cm), a second equilateral triangle inscribed by joining — Mathematics question — Figure

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Concept understanding — Geometric Progression

Geometric Progression: The Idea of Repeated Multiplication

Imagine you're folding a piece of paper in half. Start with thickness 1 unit. After one fold, thickness becomes 2. After two folds, thickness becomes 4. After three folds, thickness becomes 8. The sequence of thicknesses is:

1, 2, 4, 8, 16, ...

Notice the pattern: each term is obtained by multiplying the previous term by the same number (here, 2). That's the core intuition behind a geometric progression — you keep multiplying by a fixed number, step after step.

This is different from an arithmetic progression, where you keep adding a fixed number. Here, the growth is multiplicative, not additive. That's why geometric progressions grow (or shrink) much faster.

Precise Definition

A Geometric Progression (GP) is a sequence of numbers where the ratio of any term to its preceding term is constant. This constant is called the common ratio, denoted by $r$ .

If the first term is $a$ , then the sequence looks like:

a, \ ar, \ ar^2, \ ar^3, \ ar^4, \ \dots

Note

The common ratio $r$ can be any real number — positive, negative, or even a fraction. If $r$ is negative, the terms alternate in sign. If $0 < r < 1$ , the terms get smaller and smaller.

The $n$ -th Term

To find any term directly without listing all previous ones, use the formula:

T_n = a \cdot r^{\,n-1}

where $T_n$ is the $n$ -th term, $a$ is the first term, $r$ is the common ratio, and $n$ is the term number (starting from 1).

Example: For the paper-folding sequence, $a = 1$ , $r = 2$ . The 5th term is $1 \cdot 2^{5-1} = 2^4 = 16$ , which matches our list.

Sum of $n$ Terms

There are two cases, depending on whether $r = 1$ or not.

n

When $r = 1$ , every term is just $a$ , so the sum is simply $n \times a$ .

Why the formula works (intuition):

Let $S = a + ar + ar^2 + \dots + ar^{n-1}$ . Multiply both sides by $r$ : $rS = ar + ar^2 + \dots + ar^n$ . Subtract the first from the second: $rS - S = ar^n - a$ , so $S(r-1) = a(r^n - 1)$ , giving the formula above.

Sum of an Infinite GP

If the common ratio $r$ lies strictly between $-1$ and $1$ (i.e., $|r| < 1$ ), the terms get smaller and smaller, and the sum of all terms approaches a finite value:

S_\infty = \frac{a}{1 - r}, \quad \text{for } |r| < 1

Watch out

If $|r| \geq 1$ , the infinite sum does not exist (it diverges to infinity or oscillates without settling). Never apply the infinite sum formula when $|r| \geq 1$ .

Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ has $a=1$ , $r=\frac12$ , so $S_\infty = \frac{1}{1 - 1/2} = 2$ . This matches the intuition that repeatedly halving a unit length eventually fills exactly 2 units.

Quick Reference Table

Property	Formula	Condition
Common ratio	$r = \dfrac{T_{n+1}}{T_n}$	Always
$n$ -th term	$T_n = a r^{n-1}$	Always

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Each inscribed triangle has a side exactly half its predecessor's; the sixth inscribed triangle has perimeter $\dfrac{15}{16}$ cm.

Why this works: the geometry of midpoint triangles

Joining the midpoints of an equilateral triangle's three sides creates a new, smaller equilateral triangle inside it. By the midpoint theorem, the segment joining the midpoints of two sides is parallel to the third side and exactly half its length. So every time we repeat this construction, the new triangle's side is half the previous triangle's side — this is a geometric progression with common ratio $\dfrac{1}{2}$ .

Step-by-step

Original triangle (given): side $s_1 = 20$ cm.
1st inscribed triangle: side $s_2 = \dfrac{s_1}{2} = 10$ cm.
2nd inscribed triangle: side $s_3 = \dfrac{s_2}{2} = 5$ cm.
3rd inscribed triangle: side $s_4 = \dfrac{s_3}{2} = 2.5$ cm.
4th inscribed triangle: side $s_5 = \dfrac{s_4}{2} = 1.25$ cm.
5th inscribed triangle: side $s_6 = \dfrac{s_5}{2} = 0.625$ cm.
6th inscribed triangle: side $s_7 = \dfrac{s_6}{2} = 0.3125 = \dfrac{5}{16}$ cm.

In general, the side of the $n$ th triangle overall (counting the original as $n=1$ ) is

$s_n = 20 \cdot \left(\dfrac{1}{2}\right)^{n-1}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 43 on this concept.

KEAM 2026Set eng-2026-04174 marksMCQ
$r>0$
›Reveal solutionSolution

From $a+ar=12$ and $ar^2=16$ derive $3r^2-4r-4=0$ ; the positive root is $r=2$ .

Let the GP have first term $a$ and ratio $r$ . Then
$a(1+r)=12,\qquad ar^2=16.$
From the first, $a=\dfrac{12}{1+r}$ . Substituting:

$\frac{12r^2}{1+r}=16\Rightarrow 12r^2=16+16r\Rightarrow 3r^2-4r-4=0.$ …
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KEAM 2026Set eng-2026-04174 marksMCQ
$Q. The sum of four consecutive terms in a geometric progression is 960. If the fourth term is 8 times as large as the first term, then the smallest number in the geometric progression is (A) 40 (B) 64 (C) 128 (D) 160 (E) 240$
›Reveal solutionSolution

The ratio is $r=2$ (since term 4 $=8\times$ term 1), and $15a=960$ gives smallest term $a=64$ .

Let the four consecutive terms be $a,ar,ar^2,ar^3$ . The condition $ar^3=8a$ gives $r^3=8$ , so $r=2$ .

The sum is

$a(1+r+r^2+r^3)=a(1+2+4+8)=15a=960\Rightarrow a=64.$ …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\frac{1}{3}$
›Reveal solutionSolution

Solve $a\cdot\tfrac{40}{27}=200$ to get first term $135$ .

Geometric series with $r=\tfrac13$ :

$S_4=a\dfrac{1-r^4}{1-r}=a\dfrac{1-\tfrac1{81}}{1-\tfrac13}=a\cdot\dfrac{80/81}{2/3}=a\cdot\dfrac{40}{27}.$ …
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KEAM 2026Set eng-2026-04184 marksMCQ
$r\neq 1$
›Reveal solutionSolution

Ratio of $t_4$ to $(t_2+t_3)$ yields $2r^2-r-1=0$ , so $r=-\tfrac12$ .

Let terms be $ar,ar^2,ar^3$ . Given:
$ar+ar^2=8,\qquad ar^3=4.$
Divide the second by the first:

$\dfrac{ar^3}{ar(1+r)}=\dfrac{r^2}{1+r}=\dfrac{4}{8}=\dfrac12.$ …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\dfrac{1}{1024}$
›Reveal solutionSolution

$r^{19}=2^{-19}$ , so $r=\tfrac12$ .

With first term $a=512=2^9$ and $t_{20}=a r^{19}$ :

$2^9\,r^{19}=\dfrac{1}{1024}=2^{-10}\Rightarrow r^{19}=2^{-19}.$ …
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KEAM 2026Set eng-2026-04184 marksMCQ
$k$
›Reveal solutionSolution

Solve $k^2+5k-36=0$ ( $k=4,-9$ ), then $r=6/k=\tfrac32$ or $-\tfrac23$ .

For a GP, the middle term squared equals the product of neighbours:

$6^2=k(k+5)\Rightarrow k^2+5k-36=0\Rightarrow(k-4)(k+9)=0.$ …
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KEAM 2026Set eng-2026-04194 marksMCQ
$2, 6, 18, \ldots, 1458$
›Reveal solutionSolution

Geometric sequence with ratio 3; solve $2\cdot3^{k-1}=1458$ .

The sequence $2, 6, 18, \dots$ is a GP with first term $a=2$ and ratio $r=3$ .

The $k$ -th term is $t_k = 2\cdot 3^{k-1}$ . Set $2\cdot3^{k-1} = 1458$ : …
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KEAM 2026Set eng-2026-04194 marksMCQ
$t_1, t_2, t_3, \ldots, t_{2n - 2}, t_{2n - 1}, t_{2n}$
›Reveal solutionSolution

Every-other term of a GP forms a GP with ratio $r^2$ .

In a GP $t_k=t_1 r^{k-1}$ . The odd-indexed subsequence $t_1,t_3,t_5,\dots$ has

$\dfrac{t_{k+2}}{t_k}=\dfrac{t_1 r^{k+1}}{t_1 r^{k-1}}=r^2.$ …
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KEAM 2026Set eng-2026-04194 marksMCQ
$\frac{4^{n + 1} + 16^{n + 1}}{4^n + 16^n}$
›Reveal solutionSolution

The GM of 4 and 16 is 8; solving yields $n=-\tfrac12$ .

Geometric mean of 4 and 16 is $\sqrt{4\cdot16}=8$ . So
$\dfrac{4^{n+1}+16^{n+1}}{4^n+16^n}=\dfrac{4\cdot4^n+16\cdot16^n}{4^n+16^n}=8.$
Let $a=4^n$ , $b=16^n$ : $4a+16b=8(a+b)\Rightarrow8b=4a\Rightarrow a=2b$ , i.e. $4^n=2\cdot16^n$ . …
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KEAM 2026Set eng-2026-04194 marksMCQ
$\frac{1}{2}$
›Reveal solutionSolution

Solve $\dfrac{448r-7}{r-1}=889$ to get $r=2$ .

For a GP with first term $a=7$ , last term $l=448$ , sum $S=\dfrac{lr-a}{r-1}$ :

$\dfrac{448r-7}{r-1}=889.$ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$t_1, t_2, t_3, \ldots, t_n$
›Reveal solutionSolution
For a geometric progression the ratio $t_m/t_n=r^{m-n}$ . The given quantity is $(r^2)^3=r^6$ ; among the options only $(t_7/t_4)^2=(r^3)^2=r^6$ matches.

Let the common ratio be $r$ , so $t_k=t_1 r^{k-1}$ and $t_m/t_n=r^{m-n}$ .

Given quantity:
$\left(\frac{t_4}{t_2}\right)^3=(r^{4-2})^3=(r^2)^3=r^6.$
Evaluate each option:

(A) $(t_7/t_3)^2=(r^4)^2=r^8$

(B) $(t_7/t_4)^2=(r^3)^2=r^6$ ✓ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$n$
›Reveal solutionSolution

A G.P. sum has the form $S_n=\frac{a}{1-r}(1-r^n)=\frac{a}{1-r}-\frac{a}{1-r}r^n$ . Writing the given $S_n=100-100^{1-n}=100-100\cdot(1/100)^n$ shows the geometric factor is $r^n=(1/100)^n$ , so $r=1/100$ .

For a G.P. with first term $a$ and ratio $r$ ,
$S_n=\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}-\frac{a}{1-r}\,r^n.$
The given sum is

$S_n=100-100^{1-n}=100-100\cdot 100^{-n}=100-100\left(\tfrac{1}{100}\right)^n.$ …
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Q.A side of an equilateral triangle is 202020 cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.

Why this works: the geometry of midpoint triangles

Step-by-step

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Q.A side of an equilateral triangle is $20$ cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued as shown in the accompanying diagram. Find the perimeter of the sixth inscribed equilateral triangle.

Unlock everything free for 14 days