CBSE Sample Papers for Class 11 Maths Set 5 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 5 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 5 with Solutions

Time Allowed: 3 hours
Maximum Marks: 80

General Instructions:

All questions are compulsory.
The question paper consists of 36 questions divided into 4 Sections A, B, C and D.
Section A comprises of 20 questions of 1 mark each, Section B comprises of 6 questions of 2 marks each, Section C comprises of 6 questions of 4 marks each and Section D comprises of 4 questions of 6 marks each.
There is no overall choice. However internal choice has been provided in 6 questions of 1 mark, 2 questions of 2 marks, 2 questions of 4 marks and 2 questions of 6 marks. You have to attempt only one of the alternatives in all such questions.
Write the serial number of questions before attempting.
Use of a calculator is not permitted.

Section – A
Question numbers 1 to 10 carries 1 mark each. For each of these questions, four alternative choices have been provided of which only one is correct. Select the correct choice:

Question 1.
Consider the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. If 1 is added to each number the variance of the numbers, so obtained is MO
(A) 6.5
(B) 2.87
(C) 3.87
(D) 8.25
Answer:
(D) 8.25

Explanation:
We know that the variance of first ten natural numbers is 8.25.

Question 2.
If without repetition of the digits four-digit numbers are formed with the numbers 0, 2, 3 and 5 then the probability of such a number divisible by 5 is
(A) \(\frac{1}{5}\)
(B) \(\frac{4}{5}\)
(C) \(\frac{1}{30}\)
(D) \(\frac{5}{9}\)
Answer:
(D) \(\frac{5}{9}\)

Explanation:
We have,
Probability of number divisible by 5 is
= No. of favorable outcomes/Total outcomes
= \(\frac{5}{9}\)

Question 3.
The standard deviation of some temperature data in °C is 5. If the data were converted into °F, then the variance would be
(A) 81
(B) 57
(C) 36
(D) 25
Answer:
(A) 81

Explanation:
We have,
SD in C = 5
Using F = 9/5 C + 32
We get the Variance in F = 81

Question 4.
x and b are real numbers. If b > 0 and | x | > b, then
(A) x ∈ (-b, ∞)
(B) x ∈ (-∞,b)
(C) x ∈ (-b,b)
(D) x ∈ (-∞, -b) ∪ (b, ∞)
OR
What will be the required interval for given number line?
(A) x ∈ (-∞, – 2)
(B) x ∈ (-∞, -2]
(C) x ∈ (-2, ∞]
(D) x ∈ [-2, ∞)

Answer:
(D) x ∈ (-∞, -b) ∪ (b, ∞)

Explanation:
We have,
b > 0 and |x| > b
so x ∈ (-∞, -b) ∪ (b, ∞)

Explanation:
We know that,
From the given number line,
We get the required interval as
x ∈ (-2, ∞]

Question 5.
If the distance between the points (a, 0, 1) and (0, 1, 2) is √27 , then the value of a is
(A) 5
(B) ± 5
(C) – 5
(D) None of these
OR
The point (-2, -3, -4) lies in the
(A) first octant
(B) seventh octant
(C) second octant
(D) eight octant
Answer:
(B) ± 5

Explanation:
We have,
Using distance formula,
⇒ a² + 1 + 1 = 27
⇒ a² = 25
⇒ a = ± 5

(B) seventh octant

Explanation:
We know that,
All the coordinates are negative in seventh octant in 3D geometry.

Question 6.
If f(x) = x¹⁰⁰ + x⁹⁹ + ……. + x + 1, then f(1) is equal to :
(A) 5050
(B) 5049
(C) 5051
(D) 50051
Answer:
(A) 5050

Explanation:
We have,
f(x) = x¹⁰⁰ + x⁹⁹ + ………….. + x + 1
f(x) = \(\frac{x^{100-1}}{x-1}\)
f(1) = 5050

Question 7.
The distance between the foci of a hyperbola is 16 and its eccentricity is √2 . Its equation is
(A) x² – y² = 32
(B) \(\frac{x^2}{4}-\frac{y^2}{9}\) = 1
(C) 2x – 3y² = 7
(D) None of these
Answer:
(A) x² – y² = 32

Explanation:
We know that
Required equation of hyperbola with focal distance 16 and eccentricity √2 is
x² – y² = 32

Question 8.
If in an A.E S_n = qn² and S_m = qm², where S_r denotes the sum of r terms of the A.P. then S_q equals
(A) \(\frac{q^3}{2}\)
(B) mnq
(C) q³
(D) (m + n)q²
Answer:
(C) q³

Explanation:
We have
S_n = qn²
S_m = qm²
So we have,
S_q = q × q² = q³

Question 9.
The distance of the point of intersection of the lines 2x – 3y + 5 = 0 and 3x + 4y = 0 from the line 5x – 2y = 0
(A) \(\frac{130}{17 \sqrt{29}}\)
(B) \(\frac{130}{7 \sqrt{29}}\)
(C) \(\frac{130}{7}\)
(D) None of these
Answer:
(A) \(\frac{130}{17 \sqrt{29}}\)

Explanation:
We have,
Point of intersection of lines as \(\left(-\frac{20}{17}, \frac{15}{17}\right)\)
Distance of point from line 5x – 2y = 0 is
= \(\frac{130}{17 \sqrt{29}}\)

Question 10.
The real value of 0 for which the expression \(\) is a real number is :
(A) nπ + \(\frac{\pi}{4}\)
(B) nπ + (- 1)ⁿ \(\frac{\pi}{4}\)
(C) 2nπ ± \(\frac{\pi}{2}\)
(D) none of these
OR
If \(\left(\frac{1+i}{1-i}\right)^x\) = 1, then
(A) x = 2n + 1
(B) x = 4n
(C) x = 2n
(D) x = 4n + 1, where n ∈ N
Answer:
(C) 2nπ ± \(\frac{\pi}{2}\)

Explanation:
We have,
\(\frac{17 i \cos x}{1-2 i \cos x}\) as real
If x is 2nπ ± \(\frac{\pi}{2}\)

(B) x = 4n

Explanation:
We have,
\(\frac{1+i}{1-i}\) = i
⇒ i^x = 1
⇒ i^x = 1
⇒ (i)^x = (i)^4x = 1
⇒ x = 4n [n is natural number]

Question, numbers 11 to 15 carry 1 mark each.
Write whether the statement is true/false.

Question 11.
If | x | < 4, then x 6 [-4,4]
Answer:
True

Explanation:
If | x |< 4 then x belongs to [-4, 4]
So given statement is true.

Question 12.
The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34).
Answer:
True

Explanation:
It is true that the equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

Question 13.
The probabilities that a typist will make 0,1,2,4 and 5 or more mistake in typing are respectively, 0.12, 0.25, 0.36, 0.14, 0.08 and 0.11.
Answer:
False

Explanation:
As the given probabilities are not possible for a given typist simultaneously for same experiment
Hence given statement is False.

Question 14.
One value of 0 which satisfies the equation sin⁴ θ – 2 sin² θ – 1 lies between 0 and 2π.
Answer:
False

Explanation:
We have,
sin⁴ x – 2 sin² x – 1 = 0
All the solutions of given equation do not lie between 0 and 2π.
So it is False.

Question 15.
The line lx + my + n = 0 will touch the parabola y² = 4ax, if ln = am²
Answer:
True

Explanation:
It is true that the line lx + my + n = 0 will touch the parabola y² = 4ax, if ln = am²

Question numbers 16 to 20 carry 1 mark each.

Question 16.
Find the coefficient of x in the expansion of (1 – 3x + 7x²)(1 – x)¹⁶.
OR
Expand \(\left(\sqrt{\frac{x}{a}}-\sqrt{\frac{a}{x}}\right)^6\) using binomial theorem.
Answer:
Given, expansion = (1 – 3x + 7x²) (1 – x)¹⁶
= (1 – 3x + 7x²)(¹⁶C₀ 1¹⁶ – ¹⁶C₁ 1¹⁵ x¹ + ¹⁶C₂ 1¹⁴ x² + ……. + ¹⁶C₁₆x¹⁶)
= (1 -3x + 7x²) (1 – 16x + 120x² + …)
∴ Coefficient of x = – 3 – 16 = – 19

Commonly Made Error:
Some students write the number of terms as n – 1 which leads to incorrect answer.

Answering Tip:
Total number of terms in expansion of (a + b)ⁿ is n + 1.

Question 17.
Find the centre and radius of the circle whose equation is 3x² + 3y² + 6x – 4y – 1 = 0
OR
Find the eccentricity of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}\) = 1.
Answer:
The given equation of the circle is
3x² + 3y² + 6x – 4y – 1 = 0

Given equation of the ellipse is

Question 18.
If P = {1, 3}, Q = {2, 3, 5}, find the number of relations from P to Q.
Answer:
Given, P = {1, 3} and Q = {2, 3, 5}
∵ n(P) = 2 and n(Q) = 3
Number of relations = 2^{n(p) × n(Q)}
= 2^{2 × 3}
= 2⁶
= 64

Commonly Made Error:
Students use the incorrect formula for finding the number of relations.

Answering Tips:
Learn the formulae of the number of relations of two sets,
Number of relations = 2^{n(p) × n(Q)}
Practice more such problems based on this formula.

Question 19.
How many different words can be formed by using all the letters of word ‘SCHOOL’ ?
OR
Ten buses are plying between two places A and B. In how many ways a person can travel from A to B and come back?
Answer:
Since ‘SCHOOL’ has 6 letters, out of these 6 letters there is 2 O’s.
Hence, the number of permutations
= \(\frac{6 !}{2 !}=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2}\)
= 360

Commonly Made Error:
Students don’t know about the method of ! finding number of permutations for repeated terms.

Answering Tip:
Students should learn all the methods for finding the number of permutations in all cases.

Given, there are 10 buses from A to B.
So, a person can select any out of 10, therefore, there are 10 ways to travel from A to B. Similarly, he can come back in 10 ways.
Therefore, total number of ways to travel from A to B round trip = 10 × 10 = 100 ways

Question 20.
Show that the ∆ABC with vertices A (0, 4, 1), B(2, 3, -1) and C (4, 5, 0) is right-angled.
Answer:
Given that, the vertices of the ∆ABC are A (0,4,1), B (2, 3,-1) and C (4, 5, 0).

It satisfies the Pythagoras Therorem so the given vertices form a right angled triangle.
∵ AC² = AB² + BC²
⇒ 18 = 9 + 9
Hence, ∆ABC is a right angled triangle.

Section – B
Question numbers 21 to 26 carry 2 marks each.

Question 21.
Evaluate

Answer:

Question 22.
Find the maximum and minimum value of 7 cos x + 24 sin x.
OR
Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 3 = 0 may intersect at one point.
Answer:
y = 7 cos x + 24 sin x
y² = (7 cos x + 24 sin x)²
= 49 cos² x + 576 sin² x + 2 × 7 × 24 cos x sin x
= 49 – 49 sin² x + 576 – 576 cos² x + 2 × 7 × 24 cos x sin x
= 625 – (7 sin x – 24 cos x)²
For Maximum value
7 sin x – 24 cos x = 0
7 sin x = 24 cos x

∴ Maximum value = 25

The equation of the given lines are
3x + y – 2 = 0 …(1)
px + 2y- 3 =0 …(2)
2x -y – 3 = 0 …(3)
On solving equations (1) and (3), we obtain x = 1 and y = -1
Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).
p (1) + 2(- 1) – 3 = 0
p – 2 – 3 = 0
p = 5
Thus, the required value of p is 5.

Question 23.
If E₁, E₂, E₃ are three mutually exclusive event and less space between words of an experiment such that
2P(E₁) = 3P(E₂) = P(E₃)/ then find P(E₁).
Answer:
Since E₁, E₂, E₃ are mutually exclusive and exhaustive events, so E₁ ∩ E₂ = Φ, E₂ ∩ E₃ = Φ, E₁ ∩ E₃ = Φ, E₁ ∩ E₃ ∩ E₃ = Φ and E₁ ∪ E₂ ∪ E₃ = Φ
∴ p(E₁ ∪ E₂ ∪ E₃) = p(E₃) + p(E₂) + p(E₃)
⇒ P(S) = P(E₁) + \(\frac{2}{3}\)P(E₁) + 2P(E₁)
⇒ 1 = P(E₁) + \(\frac{8}{3}\)P(E₁)
⇒ \(\frac{11}{3}\)P(E₁) = 1
⇒ P(E₁) = \(\frac{3}{11}\)

Question 24.
Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ .
OR
Using properties of sets, show that A ∪ (A ∩ B) = A
Answer:
Let take three sets A = {1, 2}, B = {2, 3} and C = {3, 1 }; A ∩ B, B ∩ C and A ∩ C should be non empty sets. A ∩ B = {2}, B ∩ C = {3} and A ∩ C = {1}
Therefore, A ∩ B, B ∩ C and A ∩ C are non empty. Intersection of all three sets is null set, A ∩ B ∩ C = Φ
OR
A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) = A ∩ (A ∪ B) = A

Question 25.
Find the equation of the lines which cut-off intercepts on the axes whose sum and products are 1 and – 6 respectively.
Answer:
Equation of a line in intercept form is
\(\frac{x}{a}+\frac{y}{b}\) = 1
Given, a+b = 1
and a × b = – 6
⇒ a – \(\frac{6}{a}\) = 1
⇒ a² – a – 6 = 0
⇒ (a – 3) (a + 2) = 0
⇒ a = 3, -2.
If a = 3, b = – 2 and if a = – 2, b = 3.
∴ Equation of the line is
\(\frac{x}{3}-\frac{y}{2}\) = 1 or \(\frac{x}{-2}+\frac{y}{3}\) = 1
⇒ 2x – 3y – 6 = 0 or 3x – 2y + 6 = 0
2x – 3y – 6 = 0 or 3x – 2y + 6 = 0, is the required equation of the line.

Question 26.
Evaluate

Answer:

Section – C
Question numbers 27 to 32 carry 4 marks each.

Question 27.
If α and β are the solution of the equation, a tan θ + b sec θ = c, then show that tan (α + β) = \(\frac{2 a c}{a^2-c^2}\)
OR
Prove that: \(\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}=-\frac{1}{16}\)
Answer:
Given, a tan θ + b sec θ = c
⇒ (a tan θ – c)² = b² (1 + tan ²θ)
⇒ (a² tan² θ) – 2ac tan θ + c² = b² + b² tan² θ
⇒ (a² – b²) tan²θ – 2ac tan θ + c² – b² = 0
Since, α and β are roots of the equation (i),
we have,

Question 28.
If in the expansion of (1 + x)ⁿ, the coefficients of three consecutive terms are 28, 56 and 70. Then find n and the position of terms of these coefficients.
Answer:
Let coefficient of terms be ⁿC_r, ⁿC_{r + 1}, ⁿC_r+2.
Now, \(\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{n-r}{r+1}=\frac{56}{28}\) = 2
After cross multiplying, we get
n – r = 2(r + 1)
⇒ n = 3r + 2
⇒ 4n = 12r + 8
Now, \(\frac{{ }^n C_{r+2}}{{ }^n C_{r+1}}=\frac{70}{56} \Rightarrow \frac{n-r-1}{r+2}=\frac{5}{4}\)
After cross multiplying, we get
4(n – r – 1) = 5(r + 2)
⇒ 4n = 9r + 14
Equating (i) and (ii), we get
9r + 14 = 12r + 8
⇒ 3r = 6 or r = 2
Substituting the value of r in (i), we get in
4n = 12 × 2 + 8 or 4n = 32 or n = 8
Position of 3 consecutive terms are 3, 4 and 5. i.e., 3^rd term, 4^th term and 5^th term, respectively.

Question 29.
Find the real value of x and y if \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-i}\) = i
Answer:
We have,

4x + 9y – 3 = 0 and 2x – 7y – 13 = 0
x = 3 and y = – 1 are the required real values.

Question 30.
Find the value of ‘k’ if

OR
Find the mean and variance for the first 10 multiples of 3.
Answer:
Given,

Commonly Made Error:
Some students make arithmetic errors in their work when solving a limit for an unknown value.

Answering Tip:
Students need to revise standard limits formula and practice more such problems based on it.

First 10 multiples of 3 are:
3, 6, 9, 12, 15, 18, 21, 24, 27, 30

x_i	y_i= \( \frac{x_i-15}{3} \)	y_i²
3	– 4	16
6	-3	9
9	-2	4
12	-1	1
15	-0	0
18	1	1
21	2	4
24	3	9
27	4	10
30	5	25
Total	5	85

Mean, x̄ = a + \(\frac{\sum y_i}{n}\) × h
= 15 + \(\frac{5}{10}\) × 3
= 15 + 1.5
= 16.5

Variance, σ² = \(\frac{h^2}{n^2}\) [nΣyi² – (Σyi)²]
= \(\frac{9}{100}\) (850 – 25)
= 74.5
∴ Variance = 74.25 and mean = 16.5

Question 31.
Find the co-ordinates of focus, the axis of the parabola, the equation of directrix, co-ordinate of vertex of the parabola x² – 6x – 4y – 11 = 0
Answer:
Given equation of the parabola is
x² – 6x – 4y – 11 = 0
⇒ (x² – 6x + 9) – 9 – 4y – 11 = 0
⇒ (x – 3)² = 4y + 20
⇒ (x – 3)² = 4(y + 5) = 4 . 1 . (y + 5)
For vertex, x – 3 = 0 ⇒ x = 3
y – 5 = 0 ⇒ y = -5 1
∴ vertex (3, – 5)
Axis of parabola x – 3 = 0 ⇒ x = 3
For directrix, y + 5 = -1
⇒ y + 6 = 0 .
And for focus, y + 5 = 1 and x – 3 = 0
⇒ y = -4
⇒ x = 3
⇒ focus = (3, -4), axis of parabola is x = 3, equation of directrix is y + 6 = 0 and co-ordinates of vertex are (3, -5).

Question 32.
The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7,10,
12,12 and 13, find the remaining two observations. STD
Answer:
Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.
⇒ 60 + x + y = 72
⇒ x + y = 12

⇒ x² + y² = 80
From (1), we obtain ……. (2)
x² + y² + 2xy = 144 ………. (3)
From (2) and (3), we obtain
2xy = 64 …….. (4)
Subtracting (4) from (2), we obtain
x² + y² – 2xy = 80 – 64 = 16
⇒ x – y = ± 4 …… (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 4, when x – y = 4
x = 4 and y = 8, when x – y = – 4 2
Thus, the remaining observations are 4 and 8.

Section – D
Question numbers 33 to 36 carry 6 marks each.

Question 33.
In a survey of 60 people it was found that 25 people read newspaper H, 26 read newspaper T, 26 read the
newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all the three newspapers find: Up]
(i) The number of people who read at least one newspaper.
(ii) The number of people who read exactly one newspaper.
Answer:
Given
n (U) = 60, n(H ∩ I) = 9
n(H) = 25, n(H ∩ T) = 11
n(T) = 26, n(T ∩ I) = 8
n (I) = 26, n(H ∩ T ∩ I) = 3

(i) The number of people who read at least one newspaper = n(H ∪ T ∪ I)
n(H ∪ T ∪ I) = n(H) + n(T) + n(I) – n(H ∩ I)
– n(H ∩ T) – n(T ∩ r) + n(H ∩ T ∩ I)
= 25 + 26 + 26 – 9 – 11 – 8 + 3
= 80 – 28
= 52
∴ The number of people who read at least one newspaper = 52

(ii) The number of people who read exactly one newspaper
n(H) + n(T) + n(l) – 2[n(H ∩ I) + n(H ∪ T) n(T ∪ I) + 3 n(H ∩ T ∩ I)
= 25 + 26 + 26 – 2(9 + 11 + 8) + 3 × 3
= 77 – 56 + 9
= 86 – 56 = 30
∴ The number of people who read exactly one newspaper = 30.

Question 34.
Find the distance of the line 4x + 7y + 5 = 0 from the point (1,2) along the line 2x – y = 0.
OR
Evaluate

Answer:
The given lines are
2x – y = 0 …… (1)
4x + 7y + 5 = 0 ……. (2)
A (1, 2) is a point on line (1).
Let B be the point intersection of line (1) and (2).

On solving equations (1) and (2), we obtain x = \(\frac{-5}{18}\) and y = \(\frac{-5}{9}\)
∴ Coordinates of point B are \(\left(\frac{-5}{18}, \frac{-5}{9}\right)\)
By using formula, the distance between points A and B can be obtained as

Thus, the required distance is \(\frac{23 \sqrt{5}}{18}\) units.

Question 35.
Solve the following system of inequalities graphically 4x + 3y ≤ 60, y ≥ 2x, x, y ≥ 0, x ≥ 3.
Answer:
The inequalities are
4x + 3y ≤ 60, y ≥ 2x, x, y ≥ 0, x ≥ 3

(a) The line 4x + 3y = 60 passes through (15, 0), (0, 20) and it is represented by AB.
Putting x = 0, y = 0, we have
0 + 0 = 0 < 60, which is true, therefore, origin lies in this region. Thus, region below AB and the points lying on AB represent the inequality 4x + 3y ≤ 60. (b) The line y ≥ 2x passes through (0, 0), (5, 10). It is represented by CD. Putting x = 0, y = 5 in y – 2x ≥ 0 5 > 0, which is true
∴ (0, 5) lies in this region.

The region lying above the line CD and including the points on CD represents y ≥ 2x.

(c) x ≥ 3 is the region lying on the right of line x = 3, which is represented EF and points lying on x = 3 represent the inequality x ≥ 3.
The shaded area is the ∆POR in which x > 0 and y > 0 is true for each point, is the solution of given inequalities.

Question 36.
The sum of three numbers in G.E is 56. If we subtract the 1,7,21 from these numbers in that order, we obtain an Arithmetic progression (A.E). Find the numbers.
OR
Find the mean and variance for the following frequency distribution.

Answer:
Let three numbers in G.E are a, ar, ar².
Given, a + ar + ar² = 56 ……….. (i)
Again, a – 1, ar – 7, ar² – 21 are in A.E
⇒ 2(ar – 7) = (a – 1) + (ar² – 21)
(∵ If a, b, c are in A.E, then 2b = a+c)
⇒ 2ar -14 = a + ar² – 22
⇒ a + ar² – 2ar = -14 + 22
⇒ a + ar² – 2ar = 8 …………… (ii)
Dividing Eq. (i) by Eq. (ii), we get
\(\frac{a+a r+a r^2}{a+a r^2-2 a r}\) = \(\frac{56}{8}\)
⇒ \(\frac{1+r+r^2}{1+r^2-2 r}\) = \(=\frac{7}{1}\)
⇒ 1 + r + r² = 7 + 7r² – 14r
⇒ 6r² – 15r + 6 = 0
Dividing by 3
⇒ 2r² – 5r + 2 = 0
Factorizing it by splitting the middle term,
⇒ 2r² – (4 + 1)r + 2 = 0
⇒ 2r² – 4r – r + 2 = 0
⇒ 2r(r – 2) – (r – 2) = 0
⇒ (r – 2) (2r – 1) = 0
⇒ r = 2, \(\frac{1}{2}\)
If r = 2, then from Eq. (i),
a + 2a + 4a = 56
⇒ 7a = 56
a = 8
Then, numbers are
a = 8
ar = 8 × 2 = 16
⇒ ar² = 8 × 4 = 32
8, 16, 32
If r = \(\frac{1}{2}\), then from Eq. (i),
\(\frac{a}{1}+\frac{a}{2}+\frac{a}{4}\) = 56
\(\frac{4 a+2 a+1}{4}\) = 56
⇒ \(\frac{7 a}{4}\) = 56
⇒ a = 32
Then, numbers are
a = 32
ar = 32 × \(\frac{1}{2}\) = 16
or ar² = 32 × \(\frac{1}{4}\) = 8
⇒ 32, 16, 8
Hence, required numbers are 8,16, 32 or 32,16, 8.