Exploring Some More Progressions Class 9 Extra Questions Advanced Maths Chapter 6

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Class 9 Exploring Some More Progressions Extra Questions

Exploring Some More Progressions Class 9 Very Short Question Answer

Question 1.
Find the sum of first 9 terms of the GP 1, 2, 4, 8, …………… .
Answer:
Given, GP is 1, 2, 4, 8, … upto 9 terms.
Here, first term, a = 1 and n = 9.
Now, common ratio, r = \(\frac{2}{1}\) = 2
∴ The sum of the first n terms of a GP,
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\), r ≠ 1
On substituting all the values, we get
S₉ = \(\frac{1\left(2^9-1\right)}{2-1}\) = \(\frac{512 – 1}{1}\) = 511
Hence, the sum of first 4 terms is 15.

Question 2.
Find the sum of the geometric series 3 + 6 + 12 + … + 1536.
Answer:
Hint Given GP is 3 + 6 + 12 + … + 1536.
Here, a = 3, r = \(\frac{6}{3}\) = 2 > 1
Now, use S_n = \(\frac{a\left(r^n-1\right)}{r-1}\), r > 1
= 3069

Question 3.
Find the sum of first 5 terms of the GP 2, 4, 8, 16, …
Answer:
Given, a = 2 and, r = \(\frac{4}{2}\) = 2
Also, n = 5
Now, the sum of first n terms of GP,
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\), r > 1
On substituting all the values, we get
S₅ = \(\frac{2\left(2^5-1\right)}{2-1}\) = 2(32 – 1) = 2 × 31 = 62
Hence, the sum of the first 5 terms is 62.

Question 4.
Find the sum of the first n terms of the GP p, q, \(\frac{q^2}{p}\), \(\frac{q^3}{p^2}\), ……. for \(\frac{q}{p}\) > 1
Answer:
Hint First term = p and common ratio, r = \(\frac{p}{q}\)

Question 5.
Find the value of S_∞ for the GP 4, 2, 1, \(\frac{1}{2}\), … .
Answer:
Given, GP is 4, 2, 1, \(\frac{1}{2}\), …
Here, a = 4 and r = \(\frac{2}{4}\) = \(\frac{1}{2}\)
Since, r < 1
Therefore, the sum of an infinite GP exists.
We know that S_∞ = \(\frac{a}{1 – r}\)
On substituting the values, we get
S_∞ = \(\frac{4}{1-\frac{1}{2}}=\frac{4}{\frac{1}{2}}\) = 8
Hence, the sum to infinity is 8.

Question 6.
Find the sum to infinity of the GP 3, 1, \(\frac{1}{3}\), \(\frac{1}{9}\), … .
Answer:
Given, a = 3 and r = \(\frac{1}{3}\)
Since, r < 1
Therefore, the sum of an infinite GP exists.
We know that S_∞ = \(\frac{a}{1 – r}\)
On substituting all the values, we get
S_∞ = \(\frac{3}{1-\frac{1}{3}}=\frac{3}{\frac{2}{3}}\) = 3 × \(\frac{3}{2}\) = \(\frac{9}{2}\)
Hence, the sum of first 4 terms is \(\frac{9}{2}\).

Question 7.
Does the GP 1, 3, 9, 27, … have a finite sum to infinity?
Answer:
Given, a = 1 and r = \(\frac{3}{1}\) = 3
Now, r = 3 > 1
Therefore, the condition for sum to infinity does not satisfy.
Hence, the GP does not have a finite sum to infinity.

Question 8.
Find the sum to infinity of the GP 2 + 1 + \(\frac{1}{2}\) + \(\frac{1}{4}\) +… .
Answer:
Given, GP is 2 + 1 + \(\frac{1}{2}\) + \(\frac{1}{4}\) + … .
Here, a = 2 and r = \(\frac{1}{2}\)
Since, r < 1
Therefore, the sum to infinity exists.
We know that S_∞ = \(\frac{a}{1 – r}\)
On substituting the values, we get
S_∞ = \(\frac{2}{1-\frac{1}{2}}=\frac{2}{\frac{1}{2}}\) = 4
Hence, the sum to infinity is 4.

Question 9.
Evaluate 3^1/2 × 3^1/4 × 3^1/8 × … ∞.
Answer:

Exploring Some More Progressions Class 9 Short Question Answer

Question 1.
Find the first term of GP, whose common ratio is \(\frac{1}{2}\) and the sum of first 6 terms of the GP is \(\frac{63}{64}\).
Answer:
Given, r = \(\frac{1}{2}\) and S₆ = \(\frac{63}{64}\)
Now, the sum of first n terms of a GP,
S_n = \(\frac{a\left(r^n-1\right)}{r-1}\), r < 1
On substituting the given values, we get

Hence, the first term of the GP is a \(\frac{1}{2}\).

Question 2.
Find the sum of the first 12 terms of the GP 0.2, 0.04, 0.008, ……. .
Answer:
Given, a = 0.2
Now, the common ratio,
r = \(\frac{0.04}{0.2}\) = 0.2
Also, n = 12
We know that the sum of first n terms of a GP,
S_n = \(\frac{a\left(1-r^n\right)}{1-r}\), r < 1
On substituting the given values, we get
S₁₂ = \(\frac{0.2\left[1-(0.2)^{12}\right]}{1-0.2}\)
= \(\frac{0.2\left[1-(0.2)^{12}\right]}{0.8}\) = \(\frac{1}{4}\) [1 – (0.2)¹²]

Question 3.
The first term and seventh term of a GP are 729 and 64, respectively. For r > 0, determine S₉.
Answer:
Hint Given, ar⁶ = 64 ⇒ r = \(\frac{2}{3}\) [∵ r > 0]

Question 4.
The ratio of the sum of the first three terms and the sum of the first six terms of a GP is 125 : 152. Find the common ratio.
Answer:
Let a be the first term and r be the common ratio of given GP.
Given, S₃ : S₆ = 125 : 152

Hence, the common ratio is \(\frac{3}{5}\).

Question 5.
Evaluate \(\sum_{k=1}^{11}\left(2+3^k\right)\).
Answer:
Hint \(\sum_{k=1}^{11}\left(2+3^k\right)=\sum_{k=1}^{11} 2+\sum_{k=1}^{11} 3^k\) = 11 × 2 + 3¹ + 3² + … + 3¹¹
= 265741

Question 6.
Evaluate \(\sum_{r=1}^n\left(3^r-2^r\right)\).
Answer:
Hint Let S_n = \(\sum_{r=1}^n\left(3^r-2^r\right)\)
⇒ S_n = (3¹ + 3² + 3³ + … upto n terms) – (2¹ + 2² + 2³ +…upto n terms)

Question 7.
Find the sum of the series
0.5 + 0.55 + 0.555 +… upto n terms.
Answer:
Hint Let S_n = 0.5 + 0.55 + 0.555 + … upto n terms
⇒ S_n = 5[0.1 + 0.11 + 0.111 + … upto n terms]
⇒ S_n = \(\frac{5}{9}\)[\(\frac{9}{10}\) + \(\frac{99}{100}\) + \(\frac{999}{1000}\) + …. upto n terms]
⇒ S_n = \(\frac{5}{9}\) [(1 + 1 + 1 + … upto n terms) – \(\frac{1}{10}\) + \(\frac{1}{10^2}\) + \(\frac{1}{10^3}\) + … upto n terms]
⇒ S_n = \(\frac{5}{9}\left[n-\frac{1}{10} \frac{\left(1-\frac{1}{10^n}\right)}{\left(1-\frac{1}{10}\right)}\right]\)
⇒ \(\frac{5}{9}\) [n – \(\frac{5}{9}\) (1 – 10^-n)]

Question 8.
In a GP, the ratio of the sum to infinity and the first term is 5 : 2. Find the common ratio.
Answer:
Let the first term of the GP be a and the common ratio be r.
Now, the sum of the infinite GP is given by
S_∞ = \(\frac{a}{1 – r}\), r < 1.
According to the question,
S_∞ a = 5 : 2
⇒ \(\frac{S_{\infty}}{a}\) = \(\frac{5}{2}\)
On substituting the value of S_∞, we get
= \(\frac{\frac{a}{1-r}}{\frac{a}{a}}\) = \(\frac{5}{2}\)
\(\frac{1}{ 1 – r}\) = \(\frac{5}{2}\)
⇒ 2 = 5(1 – r) ⇒ 2 = 5 – 5r
⇒ 5r = 5 – 2 ⇒ 5r = 3
⇒ r = \(\frac{3}{5}\)
Hence, the common ratio is \(\frac{3}{5}\).

Question 9.
A ball rebounds to (\(\frac{1}{2}\)) th of the height from which it falls. If it is dropped from a height of 40 m, find th total distance travelled before coming to rest.
Answer:
Given, the ball is first dropped from the height of 40 m.
After striking the ground, it rebounds to half of the previous height.
Therefore, the successive rebound heights are
20, 10, 5, \(\frac{5}{2}\), …… .
This forms a geometric progression with a = 20 and r = \(\frac{1}{2}\).
Since, the ball travels 40 m downward initially. After that, for each rebound, the ball travels upward as well as downward through the same distance.
Hence, the total distance travelled is [40 + 2(20 + 10 + 5 +…)] m.
Now, 20 + 10 + 5 + is an infinite GP.
Since, |r| < 1
Therefore, the sum to infinity exists.
We know that S_∞ =
On substituting the values, we get S_∞ = \(\frac{20}{1-\frac{1}{2}}\) = \(\frac{20}{\frac{1}{2}}\) = 40
Therefore, total distance travelled = 40 + 2(40) = 40 + 80 = 120 m
Hence, the total distance travelled by the ball before coming to rest is 120 m.

Question 10.
Find the nth term of the sequence 6, 14, 26, 42, 62,. .. and hence find its 8th term.
Answer:
The given sequence is 6, 14, 26, 42, 62, ….. .
It can be written as

Let the first elements of three rows be b, a and d, respectively.
Then, b = 6, a = 8 and d = 4
Now, the nth term is given by
t_n = bC(n – 1, 0) + a(n – 1, 1) + dC(n – 1, 2)
= 6 × 1 + 8(n – 1) + 4 × \(\frac{(n-1)(n-2)}{2}\)
= 6 + 8n – 8 + 2(n² – 3n + 2)
= 8n – 2 + 2n² – 6n + 4 = 2n² + 2n + 2
Now, putting n = 8, we get
t₈ = 2(8)² + 2(8) + 2
= 2(64) + 16 + 2 = 128 + 16 + 2 = 146
Hence, the 8th term of the sequence is 146.

Sets Class 9 Long Question Answer

Question 1.
Let S be the sum, P be the product and R be the sum of reciprocals of n term in a GP.
Prove that P²Rⁿ = Sⁿ.
Answer:
Let the GP be a, ar, ar², ar³, ….. , ar^{n – 1}.
Given, S = Sum of n terms
= a + ar + ar² + ar³ + ….. + ar^{n – 1}
= \(\frac{a\left(r^n-1\right)}{r-1}\) [let r > 1] …… (i)
and R = Sum of the reciprocals of n terms


= [a²rⁿr^-1]ⁿ = (a²r^{n – 1})ⁿ = a²ⁿr^{n(n – 1)}
= P² = LHS [from Eq. (iii)]

Question 2.
The sum of an infinite GP is 6 and the sum of the cubes of its terms is 27. Find the first term.
Answer:
Let the first term be a and the common ratio be r, where |r| < 1.
The GP is a, ar, ar², …….. .
Now, the sum of the infinite GP is a
\(\frac{a}{1 – r}\) = 6
Therefore, a = 6 (1 – r) ……… (i)
Also, the squares of the terms form another GP
a³, a³r³, a³r⁶, …….. .
The sum to infinity of this GP is given as 27.
So, \(\frac{a^3}{1-r^3}\) = 27
On substituting a = 6(1 – r) from Eq. (i), we get

⇒ 8 (1 – r)² = 1 + r + r²
⇒ 8(1 – 2r + r²) = 1 + r + r²
⇒ 8 – 16r + 8r² = 1 + r + r²
⇒ 7r² – 17r + 7=0
Using the quadratic formula,

Question 3.
The sum of an infinite GP is 6 and the sum of the sum of the squares of its terms is 12. Find the sum of the cubes of the terms of the GP.
Answer:
Let the first term of the GP be a and the common ratio be r, where r < 1.
Now, the sum of the infinite GP,
\(\frac{a}{1 – r}\) = 6
Therefore, a = 6 (1 – r) ………… (i)
Also, the squares of the terms form another GP
a², a²r², a²r⁴, …….. .
The sum to infinity of this GP is given as 12.
So \(\frac{a^2}{1-r^2}\) = 12
On substituting the value of a from Eq. (i),
we get \(\frac{[6(1-r)]^2}{1-r^2}\) = 12
⇒ \(\frac{36(1-r)^2}{(1-r)(1+r)}\) = 12
⇒ \(\frac{36(1-r)}{1+r}\) = 12
⇒ 3(1 – r) = 1 + r
⇒ 3 – 3r = 1+ r
⇒ 2 = 4r
⇒ r = \(\frac{1}{2}\)
Now, substituting r = \(\frac{1}{2}\) in Eq. (i), we get
a = 6(1 – \(\frac{1}{2}\)) = 6 × \(\frac{1}{2}\) = 3
Therefore, the GP is 3, \(\frac{3}{2}\), \(\frac{3}{4}\), ……. .
Now, the cubes of the terms form the GP
27, \(\frac{27}{8}\), \(\frac{27}{64}\), …………. .
Here, a = 27 and r = (\(\frac{1}{2}\))³ = \(\frac{1}{8}\)
Therefore, the sum to infinity of the cubes ,
S_∞ = \(\frac{576}{1-\frac{1}{2}}=\frac{576}{\frac{1}{2}}\)= 27 \(\frac{8}{7}\) = \(\frac{216}{7}\)
Hence, the sum of the cubes of the terms of the GP is \(\frac{216}{7}\).

Question 4.
Write the rational number corresponding to the decimal expansion \(0.3 \overline{56}\).
Answer:
Given, decimal expansion is \(0.3 \overline{56}\).
In series form, it can be written as
\(0.3 \overline{56}\) = 0.3 + 0.056 + 0.00056 + 0.0000056 + … + ∞.

Question 5.
The mid-points of the sides of a square of side 24 cm are joined to form another square. This process is repeated indefinitely. Find the sum of the areas of all the squares formed.
Answer:
The side of the first square is 24 cm.
Therefore, the area of the first square = (24)² = 576 cm²
Now, when the mid-points of the sides of a square are joined, a new square is formed inside it.
Since, the area of the new square is half of the area of the previous square.
Therefore, the areas of the successive squares form the GP 576, 288, 144, 72, … .
Here, a = 576 and r = \(\frac{288}{576}\) = \(\frac{1}{2}\) .
Since, |r| < 1
Therefore, the sum to infinity exists.
We know that S_∞ = \(\frac{a}{1 – r}\)
On substituting the values, we get
S_∞ = \(\frac{576}{1-\frac{1}{2}}=\frac{576}{\frac{1}{2}}\) = 576 × 2 = 1152
Hence, the sum of the areas of all the squares formed is 1152 cm².

Question 6.
The sum of an infinite GP is 57 and the sum of their cubes is 9747. Find the GP.
Answer:
Let a be the first term and r be the common ratio of an infinite GP.
Given, the sum of the infinite GP = 57
⇒ \(\frac{a}{1 – r}\) = 57 ………. (i)
and sum of the cubes = 9747
⇒ a³ + a³r³ + a³r⁶ + … = 9747
⇒ \(\frac{a^3}{1-r^3}\) = 9747 ………….. (ii)
On dividing the cube of Eq. (i) by Eq. (ii), we get
\(\frac{a^3}{(1-r)^3} \cdot \frac{\left(1-r^3\right)}{a^3}\) = \(\frac{(57)^3}{9747}\)
⇒ \(\frac{1-r^3}{(1-r)^3}\) = 19
⇒ \(\frac{1+r+r^2}{(1-r)^2}\) = 19 [∵ a³ – b³ = (a – b)(a² + b² + ab)]
⇒ 1 + r + r² = 19(1 + r² – 2r) [∵ (a – b)² = a² + b² – 2ab]
⇒ 18r² – 39r + 18 = 0
⇒ (3r – 2) (6r – 9) = 0
⇒ r = \(\frac{2}{3}\) or r = \(\frac{3}{2}\)
⇒ r = \(\frac{2}{3}\)
[∵ r ≠ \(\frac{3}{2}\) because -1 < r < 1 for an infinite GP]
On putting r = \(\frac{2}{3}\) in Eq. (i), we get
\(\frac{a}{1-\left(\frac{2}{3}\right)}\) = 57
⇒ 3a = 57
⇒ a = 19
Hence, the required GP is 19, 19 × \(\frac{2}{3}\), 19 × (\(\frac{2}{3}\))², ….
i.e. 19, \(\frac{38}{3}\), \(\frac{76}{9}\), …. .

Question 7.
After striking a floor a certain ball rebounds (\(\frac{4}{5}\))th of the height from which it falls.
Find the total distance that it travels before coming to rest, if it is gently dropped from a height of 120 m.
Answer:
Given, a ball dropped from the height of 120 m. After dropping 120 m, height after first rebound
= \(\frac{4}{5}\) × 120 =96 m
After dropping 96 m, height after second rebound
= \(\frac{4}{5}\) × 96 = 76.8 m
After dropping 76.8 m, height after third rebound
= \(\frac{4}{5}\) × 76.8 = 61.44
Continuing this way, we get a geometric series
96 + 76.8 + 61.44 + … ,
where first term, a = 96
and common ratio, r = \(\frac{4}{5}\).
The sum of an infinite geometric series is given by
S_∞ = \(\frac{a}{1 – r}\), r< 1.
So, S_∞ = \(\frac{96}{1-\frac{4}{5}}=\frac{96}{\frac{1}{5}}\) = 480
Since, after the initial drop of 120 m, each rebound covers equal upward and downward distances. Thus, total distance = 120 + 2 × 480 = 1080 m
Hence, the total distance travelled by the ball before coming to rest is 1080 m.

Question 8.
A trophy is to be made out of waste material in the form of equilateral triangle as base, on this base another equilateral triangle is kept so that its vertices are mid-point of sides of the triangle, again another equilateral triangle is kept in the second equilateral triangle obtained in the same manner and process continues. Now, sides of each equilateral triangle are decorated with green ribbon to give natural environmental look. If side of equilateral triangle at the base is 30 cm. Find the total length of ribbon required to decorate the trophy.
Answer:
Hint By the given condition, we have perimeter of first triangle = 90 cm,
perimeter of second triangle = 45 cm,
perimeter of third triangle = \(\frac{45}{2}\) cm and so on.

∴ Sum of perimeters of triangles forms an GP
90 + \(\frac{90}{2}\) + \(\frac{90}{4}\) + \(\frac{90}{8}\) … upto ∞.
Here, a = 90 and r = \(\frac{1}{2}\)
Now, use sum of an infinite GP,
S_∞ = \(\frac{a}{1 – r}\), r < 1
= 180 cm

Question 9.
Find the 8th term and the sum of the first 8 terms of the sequence 2, 7, 15, 26, 40, … .
Answer:
Let the given sequence be 2, 7, 15, 26, 40, ….. .
Now, let us form the table of differences.

We observe that the second differences are same.
If the first element in each row is denoted by b, a, d, respectively.
Then, b = 2, a = 5 and d = 3
Now, the nth term of the sequence is given by
t_n = b . ^{n – 1}C₀ + a . ^{n – 1}C₁ + d . ^{n – 1}C₂
On substituting the values, we get
t_n = 2 . ^{n – 1}C₀ + 5 . ^{n – 1}C₁ + 3 . ^{n – 1}C₂

Hence, the 8th term is 100.
Now, the sum of the first n terms is given by
S_n = b . ⁿC₁ + a . ⁿC₂ + d . ⁿC₃
On substituting the values, we get
S_n = 2 . ⁿC₁ + 5 . ⁿC₂ + 3 . ⁿC₃
Therefore,

= 16 + 140 + 168
= 324
Hence, the sum of the first 8 terms is 324.

Question 10.
Find the nth term and the sum of the first n terms of the series 5 + 15 + 32 + 58 + 95 + … .
Answer:
Let the given sequence be 5, 15, 32, 58, 95, …
Now, let us form the table of differences.

We observe that the third differences are same.
If we denote the first elements of each row by b, a, d, d₁ respectively.
Then, b = 5, a = 10, d = 7 and d₁ = 2
Now, the nth term of the sequence is given by
t_n = b . ^{n – 1}C₀ + a . ^{n – 1}C₁ + d . ^{n – 1}C₂ + d₁ . ^{n – 1}C₃
On substituting the values, we get
t_n = 5 . ^{n – 1}C₀ + 10 . ^{n – 1}C₁ + 7 . ^{n – 1}C₂ + 2 . ^{n – 1}C₃
Therefore,
t_n = 5 + 10(n – 1) + 7 . \(\frac{(n-1)(n-2)}{2}\) + 2 . \(\frac{(n-1)(n-2)(n-3)}{3 \times 2}\)
= 5 + 10(n – 1) + \(\frac{7}{2}\)(n – 1) (n – 2) + \(\frac{1}{3}\)(n – 1) (n – 2) (n – 3)
= 5 + 10n – 10 + \(\frac{7}{2}\)(n² – 3n + 2) + \(\frac{1}{3}\)(n³ – 6n² + 11n – 6)

Exploring Some More Progressions Class 9 Case Based Questions

Question 1.
A student draws a square with a side length of 20 cm. By joining the mid-points of its sides, another smaller square is formed inside it. This process is repeated infinitely, creating a sequence of nested squares, where the area of each new square is exactly half of the previous one.
(i) Identify the first term (a) and the common ratio (r) for the sequence of the areas.
(ii) (a) If the process stops after the 10th square, find the total area of the squares formed.
Or
(b) Write the specific mathematical expression for the sum of the areas of the first n squares.
(iii) If the squares are drawn infinitely, use the formula S_∞ = \(\frac{a}{1 – r}\) to calculate the total area.
Answer:
Given, the side length of the first square is 20 cm.
So, the area of the first square = 20² = 400 cm²
Now, each new square has half the area of the previous square.
Therefore, the sequence of areas forms a GP, 400, 200, 100, 50, … .
(i) Thus, the first term and common ratio are a = 400 and r = \(\frac{1}{2}\).

(ii) (a) If the process stops after the 10th square, then the finite GP sum formula used is

Or
(b) The sum of the areas of the first n squares,

(iii) If the squares are drawn infinitely, then we use
S_∞ = \(\frac{a}{1 -r}\), r < 1.
On substituting a = 400 and r = \(\frac{1}{2}\), we get
S_∞ = \(\frac{400}{1-\frac{1}{2}}=\frac{400}{\frac{1}{2}}\) = 800
Hence, the total area of all the infinitely many squares is 800 cm².

Question 2.
A ball is dropped from a height of 27m. On every bounce, it reaches a height exactly \(\frac{2}{3}\) of its previous height.
(i) Calculate the height of the first bounce and the second bounce in digits.
(ii) What is the total vertical distance the ball travels (only counting the upward bounces) during the first 3 bounces?
(iii) (a) If we only look at the downward drops starting from 27 m, what is the infinite sum of all these downward distances?
Or
(b) If the ball was dropped from a height, where the first bounce was 18 m and the total distance of all bounces (to infinity) was 54 m, what would be the common ratio (r)?
Answer:
Given, a ball is dropped from a height of 27 m.
On every bounce, it rises to \(\frac{2}{3}\) of its previous height.
Thus, the heights of successive bounces form a geometric progression with a = 27 and r = \(\frac{2}{3}\).
(i) First Bounce 27 × \(\frac{2}{3}\) = 18
Therefore, the height of the first bounce is 18 m.
Second Bounce
18 × \(\frac{2}{3}\) = 12
Therefore, the height of the second bounce is 12 cm.

(ii) The heights of the upward bounces are 18m, 12m, 8m.
Now, 8 = 12 × \(\frac{2}{3}\)
Therefore, the total upward distance travelled during the first 3 bounces is 18 + 12 + 8 = 38 m.
Hence, the required total upward distance is 38m.

(iii) (a) The downward distances are 27m, 18m, 12m, 8m, … .
This is an infinite GP with a = 27 and r = \(\frac{2}{3}\).
Since, |r|< 1, the sum to infinity exists.
Using the formula, S_∞ = \(\frac{a}{1 – r}\)
S_∞ = \(\frac{27}{1-\frac{2}{3}}=\frac{27}{\frac{1}{3}}\) = 27 × 3 = 81
Hence, the infinite sum of all downward distances is 81 m.
Or
(b) Given, first bounce height = 18 m
and total distance of all bounces to infinity = 54 m
The bounce heights form the GP
18, 18r, 18r², …… .
Therefore, S_∞ = \(\frac{18}{1 – r}\)
Given that, \(\frac{18}{1 – r}\) = 54
Now, 18 = 54 (1 – r)
⇒ 18 = 54 – 54r
⇒ 54r = 36
⇒ r = \(\frac{36}{54}\) ⇒ r = \(\frac{2}{3}\)

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