CBSE Sample Papers for Class 11 Maths Set 7 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 11 Maths with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Maths Set 7 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.
The mean deviation of the data 3, 10, 4, 7, 10, 5 from the mean is
(A) 2
(B) 2.57
(C) 3
(D) 3.75
Answer:
(B) 2.57

Explanation:
Mean deviation about the mean of the data:
3, 10, 4, 7, 10, 5
Mean deviation = 15.42/6 = 2.57

Question 2.
If seven persons are to be seated in a row. Then, the probability that two particular persons sit next to each other is m
(A) \(\frac{1}{3}\)
(B) \(\frac{1}{6}\)
(C) \(\frac{2}{7}\)
(D) \(\frac{1}{2}\)
OR
The number of different four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is
(A) 120
(B) 96
(C) 24
(D) 100
Answer:
(C) \(\frac{2}{7}\)

Explanation:
Given that,
Total number of persons = 7
Probability that two persons sit next to each other
= 7

OR

(C) 24

Explanation:
We know that,
Number of 4 digit numbers that can be formed = 4! = 24

Question 3.
The standard deviation of data 6, 5, 9, 13, 12, 8 and 10 is
(A) \(\sqrt{\frac{52}{7}}\)
(B) \(\frac{52}{7}\)
(C) √6
(D) 2
Answer:
(A) \(\sqrt{\frac{52}{7}}\)

Explanation:
We know that,
For data 6, 5, 9, 13, 12, 8, 10
The standard deviation is \(\sqrt{\frac{52}{7}}\)

Question 4.
Find f'(0) if f(x) = sin x.
(A) 10
(B) 1
(C) 2
(D) 3
Answer:
(B) 1

Explanation:
Given that,
f(x) = sin x
f ‘(x) = cos x
⇒ f ‘(0) = cos 0 = 1

Question 5.
If x < 5, then
(A) – x < – 5
(B) – x < – 5 (C) – x > – 5
(D) – x > – 5
Answer:
(C) – x > – 5

Explanation:
We know that,
If x < 5 then – x > – 5

Question 6.
Let S = {x | x is a positive multiple of 3 less than 100} P = {x | x is a prime number less than 20}. Then n(S) + n(P) is
(A) 41
(B) 31
(C) 33
(D) 30
Answer:
(A) 41

Explanation:
We have,
n(S) = 33
n(P) = 8
So n(S) + n(P) = 41

Question 7.
The total number of terms in the expansion of (x + a)¹⁰⁰ + (x – a)¹⁰⁰ after simplification is
(A) 50
(B) 202
(C) 51
(D) None of these
OR
The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is
(A) ¹⁶C_n
(B) ¹⁶C₅
(C) ¹⁶C₉
(D) ²⁰C₉
Answer:
(C) 51

Explanation:
The total number of terms after simplification will be 51.

OR

(C) ¹⁶C₉

Explanation:
The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is C(16, 9).

Question 8.
If the line \(\frac{x}{a}+\frac{y}{b}\) = 1 passes through the points (21, – 3) and (4, – 5), then (a, b) is.
(A) (1, 1)
(B) (- 1, 1)
(C) (1, -1)
(D) (- 1, – 1)
OR
A point equidistant from the lines 4x + 3y +10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is
(A) (1, -1)
(B) (1, 1)
(C) (0, 0)
(D) (0, 1)
Answer:
(D) (- 1, – 1)

Explanation:
Given that,
Line \(\frac{x}{a}+\frac{y}{b}\) = 1 = 1 passes through the point (21, – 3) a b and (4, -5) then on comparing from equations we get,
⇒ (a, b) = (-1, -1)

OR

(C) (0, 0)

Explanation:
A point equidistant from the lines
⇒ 4x + 3y + 10 = 0,
⇒ 5x – 12y + 26 = 0 and
7x + 24y – 50 = 0
is (0, 0)

Question 9.

(A) 2
(B) \(\frac{1}{2}\)
(C) \(\frac{-1}{2}\)
(D) \(\frac{1}{4}\)
Answer:
(B) \(\frac{1}{2}\)

Explanation:
On simplifying the limits

Question 10.
L is the foot of the perpendicular drawn from a point (3,4, 5) on X-axis. The coordinates of L are
(A) (3,0,0)
(B) (0,4,0)
(C) (0,0, 5)
(D) None of these
Answer:
(A) (3,0,0)

Explanation:
We have,
Foot of perpendicular drawn from point (3, 4, 5) on X-axis will be (3, 0, 0).

Question numbers 11 to 15 carry 1 mark each. State True or False.

Question 11.
The sum of probabilities of two students getting the distinction in their final examination is 1.2.
Answer:
True

Explanation:
The sum of probabilities of two students getting distinction is 1.2 It is true as the sum of probabilities can be greater than 1.

Question 12.
The locus of the point of intersection of lines √3x – y – 4√3k = 0 and √3kx + ky — 4√3 = 0 for different value of k is a hyperbola whose eccentricity is 2.
Answer:
True

Explanation:
It is true that for different values of k we get a hyperbola with eccentricity 2.

Question 13.
The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then, the other two sides are y – 3 = (x – 2).
Answer:
True

Explanation:
It is true that the equation of two other sides are (y – 3) =± (x – 2).

Question 14.
If (x – 2, y + 5) = \(\left(-2, \frac{1}{3}\right)\) are two equal ordered pairs, then x = 4, y = \(\frac{-14}{3}\)
Answer:
False

Explanation:
Given that,
(x – 2, y + 5) = (-2, 1/3)
⇒ On comparing,
⇒ x = 0 and y = -14/3
So the given statement is False.

Question 15.
If x < y and b < 0, then \(\frac{x}{y}<\frac{y}{b}\)
OR
Solution set of x + y ≥ 0 is

Answer:
False

Explanation:
x < y and b < 0, So, \(\frac{x}{y}>\frac{y}{b}\)
So the given statement is False.

OR

True

Explanation:
From the given graph
The solution set is x + y ≥ 1
So the given statement is true.

Question numbers 16 to 20 carry 1 marks each.

Question 16.
If ⁿP₄ : ⁿP₂ = 12, find n.
OR
How many ways are there to arrange the letters of the word “GARDEN” with the vowels in alphabetical order?
Answer:
Given, \(\frac{{ }^n P_4}{{ }^n p_2}\) = 12
⇒ ⁿp₄ = ¹²ⁿp₂

⇒ n² – 5n + 6 = 12
⇒ n² – 5n – 6 = 0
⇒ n² – 6n + n – 6 = 0
⇒ n(n – 6) + 1(n – 6) = 0
⇒ (n – 6)(n + 1) = 0
⇒ n = -1 or n = 6
∵ n cannot be negative
∴ n = 6

OR

Total number of ways in which all letters of the word GARDEN can be arranged = 6! =720
There are only two vowels in the word, A and E.
First place A at the first place, E can be occupy any of the remaining 5 places . Total arrangements 5 × 24.
When A in the second place, E can occupy any of 4 places
So Total arrangements 4 × 24.
Repeat this process until A occupies the last but one place. A cannot occupy the last place.
∴ The Total number of total arrangement is
(5 + 4 + 3 + 2 + 1) × 24
= 15 × 24 = 360

Question 17.
If in a A.E 7th term is 9 and 9th term is 7, then find 16th term.
OR
If S_n = 3n² + 2n, then write a₂.
Answer:
Let first term and common difference of A.E is a and d, respectively.
Given,
T₇ = 9 = a + (7 – 1)d ………. (i)
⇒ 9 = a + 6d
and T = 7 = a + (9 – 1)d
⇒ 7 = a + 8d
On solving equations (i) and (ii), we get
a = 15 and d = 1
Now, T₁₆ = a + (16 – 1)d
= 15 + 15(- 1)
= 0

Commonly Made Errors:
Students do not simplify the complex number before finding the conjugate which leads to incorrect result.

Answering Tips:
Learn and revise the formula for n^th term and practice more such problems based on it.

OR

We know that, a_n = S_n – S_{n – 1}
Given, S_n = 3n² + 2n
∴ S_{n – 1} = 3(n – 1)² + 2(n – 1) (∵ a_n = S_n – S_{n – 1})
∴ a_n = 3n² + 2n – {3(n – 1)² + 2(n -1)}
= 3n² + 2n – {3(n² – 2n + 1) + 2n – 2}
= 3n² + 2n – {3n² – 6n + 3 + 2n – 2}
= 3n² + 2n – {3n² – 4n + 1}
= 6n – 1
Hence, a₂ = 6(2) – 1 = 11

Question 18.
Write the conjugate of \(\frac{2-i}{(1-2 i)^2}\)
Answer:

Commonly Made Errors:
Students do not simplify the complex number before finding the conjugate which leads to incorrect result.

Answering Tips:
Write the conjugate of z after simplifying the polynomial equation in the form of complex number.

Question 19.
Write the value of 2 sin 75° sin 15° ?
Answer:
We know that,
2 sin A sin B = cos(A – B) – cos( A + B)
∴ 2 sin 75° sin 15° = cos (75° -15°) – cos (75° +15°)
= cos 60° + cos 90°
= \(\frac{1}{2}\) + 0
= \(\frac{1}{2}\)

Question 20.
Solve for x (x is a real number), 4x + 3 < 5x + 7.
Answer:
Given, 4x + 3 < 5x +7
Transposing 5x to LHS and 3 to RHS, we get
4x – 5x < 7 – 3
or – x < 4 or x > – 4
∴ When x is real, the solution is (- 4, ∞)

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

Question 21.
If tan A = \(\frac{1-\cos B}{\sin B}\), then tan 2A = ______________
OR
If tan A = \(\frac{a}{a+1}\) and tanB = \(\frac{1}{2 a+1}\), then find the value of A + B.
Answer:

2A = B
∴ tan 2A = tan B

OR

We know that,

Question 22.
Evaluate

Answer:

Commonly Made Error:
Some students have difficulty in eliminating (x – 1) from denominator due to the calculation

Answering Tip:
Students must be careful while doing calculation and recheck their solution.

Question 23.
Find all pairs of consecutive odd positive integers, both of which are smaller than 10 such that their sum is more than 11.
OR
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Answer:
Let x be the smaller of the two odd +ve integers. So, that the other integer is x + 2.
∴ x < 10
x + 2 <10
x < 10 – 2
⇒ x < 8 (i) Also, the sum of the two integers is more than 11. ∴ x + (x + 2) > 11
2x + 2 > 11
2x >11 – 2
2x > 9 or x > \(\frac{9}{2}\)
∴ If one number is 5 (odd number), then other is seven and if the one number is 7, the other number is 9.
∴ The possible pairs are (5, 7) and (7, 9).

OR

Let x be the smaller of the two +ve consecutive even integers, then the other one is x + 2.
Given, x > 5
And x + x + 2 < 23
2x + 2 < 23
If the sum is less than 23, it is also less than 24.
∴ 2x + 2 < 24
2x < 22 or x < 11
∴ Value of x may be 6, 8, 10 (even integers)
∴ The pairs may be (6, 8), (8, 10), (10, 12).

Question 24.
From a class of 40 students, in how many ways can five students be chosen for an excursion party.
Answer:
From a class of 40 students, five students can be chosen for an excursion party in ⁴⁰C₅ ways = 40!

= 658,008

Question 25.
If V = {x | x is positive factor of the number 2^{p – 1} (2p – 1), where 2p – 1 is a prime number}. Write Y in the roaster form.
Answer:
Given,
y = {x | x is a positive factor of the number 2^{P – 1} (2^P – 1), where 2^P – 1 is a prime number}
So, the factor of 2^{P – 1} are 1, 2, 2², 2³, . . ., 2^{P – 1}.
∴ y = {1, 2, 2², 2³, . . . , 2^{P – 1}}

Question 26.
In a large metropolitan area, the probabilities are 0.87, 0.36, 0.30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets ?
Answer:
Let E₁ be the event that family own colour television set and E₂ be the event that family own Black and White Television.
P(E₁) = 0.87
P(E₂) = 0.36
and P(E₁ ∩ E₂) = 0.30
We have to find the probability that a family owns either anyone or both kind of sets i.e, p(E₁ ∪ E₂).
Now, P(E₁ ∪ E₂) = P(E₁) + P(E₂) – P(E₁ ∩ E₂)
[by addition theorem]
= 0.87 + 0.36 – 0.30 = 0.93

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

Question 27.
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
For any sets A and B show that (A ∩ B) ∪ (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 ∩ B) ∪ (A – B) = A
L. H. S. = (A ∩ B) ∪ (A – B)
= (A ∩ B) ∩ (A ∩ B’) [∴ (A – B) = A ∩ B’]
= A ∩ (B ∪ B’) [By distributive law]
= A ∩ U [( B ∪ B’) = U = Universal set]
= A
= R. H. S.

Question 28.
The coefficients of the (r – 1)^th, r^th and (r + 1)^th terms in the expansion of (x + 1)ⁿ are in the ratio 1:3:5. Find n and r.
OR
Find the term independent of x in the expansion of \(\left(\frac{\sqrt{x}}{\sqrt{3}}+\frac{\sqrt{3}}{2 x^2}\right)^{10}\)
Answer:
In the expansion of (x + 1)ⁿ, the general term is
T_R+1 = ⁿC_Rx^n-R (1)^R or T_R+1 = ⁿC_Rx^{n – R}
For coefficient of (r – 1)^th term, put R = r – 2
T_R+1 = T_{(r – 2)+1} = ⁿC_{r – 2}x^{n – (r – 2)}
T_{(r – 1)} = ⁿC_{r – 2}x^{n – r + 2}
⇒ Coefficient of (r – 1)^th term = ⁿC_{r – 1}
For coefficient of rth term, put R = r – 1
⇒ T_r = T_r-1+1 = ⁿC_{r – 1}x^{n – (r – 1)}
⇒ T_r = ⁿC_{r – 1}x^{n – r + 1}
Coefficient of r^th term = ⁿC_r-1
For coefficient of (r + 1)^th term, put R = r
⇒ T_{r + 1} = ⁿC_rx^{n – r}
⇒ Coefficient of (r + 1)^th term = ⁿC_r
Now, according to question it is given that
ⁿC_r-2 : ⁿC_r-1 : ⁿC_r = 1 : 3 : 5
Taking first two parts

⇒ 5r = 3n – 3r + 3
⇒ 3n – 8r + 3 = 0 …… (ii)
Multiplying Eq. (i) by 2 and then subtracting from
Eq. (ii), we get
n – 7 = 0
n = 7
Putting the value of n in Eq. (i), we get
7 – 4r + 5 = 0
⇒ 4r = 12
r = 3

OR

Let (r + 1 )^th term be independent of x which is given by

Question 29.
Find all non-zero complex number z satisfying z conjugate = iz².
Answer:
Let z = x + iy
Given, z̄ = iz²
⇒ x – iy = i(x² – y² + 2ixy)
⇒ x – iy = i(x² – y²) – 2xy
⇒ (x + 2xy) – i(x² – y² + y) = 0 + 0i
⇒ x + 2xy = 0 ……….. (i)
and x² – y² + y = 0 …… (ii)
Now, x + 2 yx = 0
⇒ x(1 + y) = 0 ⇒ x = 0 or 1 + 2y = 0 ⇒ x = 0 or y = \(\frac{-1}{2}\)
Putting x = 0 in (ii), we get
⇒ – y² + y = 0 ⇒ y(y -1) = 0 ⇒ y = 0, y =1
Thus, we have the following pairs of values of x and y
x = 0, y = 0; x = 0, y = 1
∴ z = 0 + 0i = 0; z = 0 + 1i = i
Putting y = \(\frac{-1}{2}\) in (ii), we get
x² – y² + y = 0
⇒ x² – \(\frac{1}{4}-\frac{1}{2}\) = 0
⇒ x² – \(\frac{3}{4}\) = 0
⇒ x = \(\frac{\pm \sqrt{3}}{2}\)
Thus, we have the following pairs of values of x and y

Question 30.
Find the equation of circle concentric with circle 4x² + 4y² – 12x – 16y – 21 = 0 and half its area.
Answer:
The given equation of the circle is
4x² + 4y² – 12x – 16y – 21 = 0

Commonly Made Error:
Some students don’t have an understanding of that concentric circles have same centre and different radius.

Answering Tip:
Students need to emphasise on the conditions and properties of circle.

Question 31.
Find the equation of a circle whose centre is (3, -1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0
Answer:
Given centre of the circle is (3, – 1),

Now, OM = \(\left|\frac{6+5+18}{\sqrt{4+25}}\right|=\frac{29}{\sqrt{29}}\) = √29
In ∆OMB , OB² = OM² + MB²
∵ AB = 6 ⇒ AM = MB = \(\frac{1}{2}\), AB = 3
⇒ OB² = 29 + 9 ⇒ OB² = 38
So, the radius of circle is √38,
∴ Equation of the circle with radius r = √38 and centre (3, – 1) is
⇒ (x – 3)² +(y + 1)² = 38
⇒ x² – 6x + 9 + y² + 2y + 1 = 38
x² + y² – 6x + 2y = 28
x² + y² – 6x + 2y – 28 = 0 is the required equation of circle.

Commonly Made Error:
Some students don’t make the correct figure for the problem which leads to wrong assumptions.

Answering Tip:
Students must make a well labelled figure for making question easier.

Question 32.
Mean and standard deviation of 100 observation was found to be 40 and 10, respectively. If at the time of
calculation two observation was wrongly taken as 30 and 70 in place of 3 and 27 respectively, then find the correct standard deviation. Answer:

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

Question 33.
In a survey of 200 students of a school, it was found that 120 study Mathematics, 90 study physics and 70 study Chemistry 40 study Mathematics and Physics, 30 study Physics and Chemistry, 50 study Chemistry and Mathematics and 20 none of these subjects. Find the number of students who study all the three subjects.
Answer:
Let M, P and C denote the students studying Mathematics, Physics and Chemistry.
Also, n(U) =200, n(M) = 120, n(P) = 90, n(C) = 70, n(M ∩ P) = 40,
∩n(P ∩ C) =30, n(M ∩ C) = 50, n(M ∪P ∪ C)’ = 20
Now, the number of students who study all the three subjects i.e., n (P ∩ C ∩ M)
Now, we have
n(M ∪ P ∪ C)’ = 20
n(M ∪P ∪ C)’ = n(U) – n(M ∪P ∪ C)
20 = 200 – n(M ∪ P ∪ C)
n(M ∪ P ∪ C) = 200 – 20 = 180
n(M ∪ P ∪ C) = n(M) + n(P) + n(C) – n(M ∩ P) – n(P ∩ C)
– n(C ∩ M) + n(C ∩ M ∩ P)
180 = 120 + 90 + 70 – 40 – 30 – 50 + n(C ∩ M ∩ P)
180 = 280 – 120 + n(C ∩ M ∩ P)
180 + 120 – 280 = n(P ∩ C ∩ M)
n(P ∩ C ∩ M) = 300 – 280 = 20
So the number of students who study all the three subjects is 20.

Question 34.
How many words, with or without meaning can be made from the letters of the word ‘MONDAY/ assuming that no letter is repeated, if
(i) 4’ letters are used at a time.
(ii) all letters are used at a time.
(iii) all letters are used but first letter is a vowel.
OR
In how many ways can the letters of the word PERMUTATIONS be arranged if the :
(i) Words start with P and end with S,
(ii) Vowels are all together,
(iii) There are always 4 letters between P and S ?
Answer:
In the word MONDAY, all letters are different.
(i) Out of 6 different letters 4 letters can be selected in ⁶P₄, ways.
∴ Required number of words = ⁶P₄
= \(\frac{6 !}{(6-6) !}\) = \(\frac{6 !}{2 !}=\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}\) = 360

(ii) The word ‘MONDAY’ has 6 different letters.
Number of ways taking 6 letters at a time = ⁶P₆
∴ Required number of words = ⁶P₆
= \(\frac{6 !}{(6-6) !}\) = \(\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{0 !}\) = 720

(iii) First, we will fix the vowel.
In the word MONDAY, there are two vowels O and A.
∴ First letter can be chosen by 2 ways.
Number of ways taking 5 different letters from remaining 5 letters = ⁵ P₅
∴ Required number of words 5! 5!
= ⁵P₅ = \(\frac{5 !}{(5-5) !}=\frac{5 !}{0 !}\) = 5 × 4 × 3 × 2 × 1 = 120
Hence, total number of ways = 2 × 120 = 240

OR

(i) Letters between P and S are ERMUTATION. These 10 letters having T two times. These letters can be arranged in \(\frac{10 !}{2 !}\) = 1814400 ways.

(ii) There are 12 letters in the word PERMUTATIONS which have T two times.
Now the vowels a, e, i, o, u are taken together.
Let it be considered in one block.
The letters of vowels can be arranged in 5! Ways.
Thus there are 7 letters and 1 block of vowels with T two times.
∴ Required number of arrangements = \(\frac{8 !}{2 !}\) × 5!

(iii) There are 12 letters to be arranged in 12 places

These 12 letters are to filled in 12 places shown above. P may be filled up at place No. 1, 2, 3, 4, 5, 6, 7 and consequently S may be filled up at place No. 6, 7, 8, 9, 10, 11, 12 leaving four places in between. Now P and S may be filled up in 7 ways. Similarly, S and P may be filled in 7 ways.
Remaining ten letters having two T’s
∴ These can be arrange \(\frac{10 !}{2 !}\)
hence word PERMUTATIONS arrange when 4 letters between P and S
= 2 × 7 × \(\frac{10 !}{2 !}\)
= 14 × 1814400
= 25401600

Question 35.
Find the linear inequalities for which the shaded region in the given figure is the solution set.

Answer:
Consider the line 3x + 2y = 48, we observe that the shaded region and the origin are on the same side of the line 3x + 2y = 48 and (0, 0) satisfy the linear constraint 3x + 2y ≤ 48. So, we must have one inequation as 3x + 2y ≤ 48.

Now, consider the line x + y = 20. We find that the shaded region and the origin are on the same side of the line x + y = 20 and (0, 0) satisfy the constraints x + y < 20. So, the second inequation is x + y ≤ 20.

We also notice that the shaded region is above X-axis and is on the right side of Y-axis, so we must have x ≥ 0, y ≥ 0.

Thus, the linear inequations corresponding to the given solution set are 3x + 2y ≤ 48, x + y ≤ 20 and x ≥ 0, y ≥ 0.

Question 36.
If pth, qh and rth terms of an A.E and G.E are both a, b and c respectively, then show that a^{b – c}. b^{c – a}. c^{a – b} = 1.
OR
If A is the arithmetic mean and G₁, G₂ be two geometric mean between any two numbers, then prove that 2A = \(\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}\)
Answer:
Let A, d are the first term and common difference between A.P and x, R are the first term and common ratio of G.P, respectively.
According to the given condition,
A + (p – 1)d = a ……… (i)
A + (q – 1)d = b ……… (ii)
A + (r – 1)d = c ……… (iii)
and
a = xR^{p – 1} ……… (iv)
b = xR^{q – 1} ……… (v)
c = xR^{r – 1} ………. (vi)
On subtracting Eq. (ii) from Eq. (i). we get
d(p – 1 – q + 1) = a – b
⇒ a – b = d(p – q) ………. (vii)
On subtracting Eq. (iii) from Eq. (ii). we get
d(q – 1 – r + 1) = b – c
b – c = d(q – r) ………. (viii)
On subtracting Eq. (i) from Eq. (iii). we get
d(r – 1 – p + 1) = c – a
⇒ c – a = d(r – p) …….. (ix)
Now, we have to prove a^{b – c} b^{c – a} c^{a – b} = 1
Taking LHS = a^{b – c} b^{c – a} c^{a – b}
Using Eqs. (iv), (v), (vi) and (vii), (viii), (ix),
LHS = (xR^{p – 1})d^{(q – r)} (xR^{q – 1})d^{(r – p)} (xR^{r – 1})^d(p-q)
= x^{d(q – r) + d(r – p) + d(p – q)} R^{(p – 1)d(q – r) + (q – 1)d(r – p) + (r – 1)d(p – q)}
= x^{d(q – r + r – p + p – q)}
= R^{d(pq – pr – q + r + qr – pq – r + p + rp – rq – p + q)} = x⁰R⁰ = 1
– RHS

OR

Let the number be a and b
Then, A = \(\frac{a+b}{2}\)
⇒ 2A = a + b ………. (i)
and G₁, G₂ be geometric mean between a and b, then a, G₁, G₂, b are in G.P.
Let r be the common ratio.
Then, b = ar_{4 – 1} [∵ a_n = ar_{n – 1}]