OP Malhotra Class 11 Maths Solutions Chapter 14 Sequence and Series Ex 14(d)

Well-structured Class 11 OP Malhotra Solutions Chapter 14 Sequence and Series Ex 14(d) facilitate a deeper understanding of mathematical principles.

S Chand Class 11 ICSE Maths Solutions Chapter 14 Sequence and Series Ex 14(d)

Question 1.
Find the A.M. between :
(i) 6 and 12
(ii) 5 and 22
(iii) (cos θ + sin θ)² and (cos θ – sin θ)²
(iv) (x + y)² and (x – y)².
Solution:
(i) Let A be the A.M between 6 and 12
∴ A = \(\frac{6+12}{2}\) = 9
[if A be the A.M between a and b.
Then a, A, b are in A.P ∴ A = \(\frac{a-b}{2}\)]

(ii) Let A be the A.M between 5 and 22
∴ A = \(\frac{5+22}{2}\) = 13.5

(iii) Let A be the A.M between
(cos θ + sin θ)² and (cos θ – sin θ)²
Then A = \(\frac{(\cos \theta+\sin \theta)^2+(\cos \theta-\sin \theta)^2}{2}\)
= \(\frac{2\left(\cos ^2 \theta+\sin ^2 \theta\right)}{2}\) = \(\frac{2}{2}\) = 1

(iv) Let A be the A.M between (x + y)² and (x – y)²
∴ A = \(\frac{(x+y)^2+(x-y)^2}{2}\)
= \(\frac{2\left(x^2+y^2\right)}{2}\) = x² + y²

Question 2.
Insert :
(i) 3 arithmetic means between 5 and 17.
(ii) 4 arithmetic means between 17 and 52 .
Solution:
(i) Let A₁, A₂, A₃ be the three A.M’s between 5 and 17 .
Then 5, A₁, A₂, A₃, 17 are in A.P
∴ 17 = T₅ = 5 + 4d
⇒ 4d = 12 ⇒ d = 3
Thus A₁ = T₂ = a + d = 5 + 3 = 8
A₂ = T₃ = a + 2d = 5 + 6 = 11
A₃ = T₄ = a + 3d = 5 + 9 = 14
Hence the required three 3 A.M’s between 5 and 17 are 8, 11, 14.

(ii) Let A₁, A₂, A₃, A₄ are the 4 A.M’s between 17 and 52 .
Then 17, A₁, A₂, A₃, A₄, 52 are in A.P.
∴ 52 = T₆ = a + 5d = 17 + 5d
⇒ 5d = 35 ⇒ d = 7
Thus,
A₁ = T₂ = a + d = 17 + 7 = 24
A₂ = T₃ = a + 2d = 17 + 14 = 31
A₃ = T₄ = a + 3d = 17 + 21 = 38
A₄ = T₅ = a + 4d = 17 + 28 = 45
Hence the required 4 A.M’s between 17 and 52 are 24, 31,38 and 45.

Question 3.
Find two numbers whose product is 91 and whose A.M. is 10 .
Solution:
Let the two numbers are a and b
s.t ab = 91 …(1)
and 10 = \(\frac{a+b}{2}\) ⇒ a + b = 20 …(2)
∴ (a – b)² = (a + b)² – 4ab
= 20² – 4 × 91
= 400 – 364 = 36
⇒ a – b = ± 6

Case-I. When a – b = 6
On adding (2) and (3); we have
2a = 26 ⇒ a = 13
∴ from (3); b = 7

Case-II : When a – b = -6
On adding (3) and (4); we have
2a = 14 ⇒ a = 7
∴ from (4); b = 13
Hence the required numbers are 7 and 13 .

Question 4.
If p arithmetic means are inserted between a and b, prove that d = \(\frac{b-a}{p+1}\).
Solution:
Let A₁, A₂, ……, A_p are p A.M’s between a and b
Then a, A₁, A₂, ……., A_p, b are in A.P.
Let d be their common difference.
b = T_p+2 = a + (p + 2 – 1)d
⇒ b – a = (p + 1) d ⇒ d = \(\frac{b-a}{p+1}\)

Question 5.
The sum of three numbers in A.P. is 33, and the sum of their squares is 461 . Find the numbers.
Solution:
Let the three numbers in A.P are
a – d, a, a + d
Since sum of three numbers is 33 .
∴ a – d + a + a + d = 33
⇒ a = 11
and Sum of their squares is 461 .
(a – d)² + a² + (a + d)² = 461
⇒ (11 – d)² + 11² + (11 + d)² = 461
⇒ d² – 22d + 121 + 121 + d² + 121 + 22d = 461
⇒ 2d² + 363 = 461 ⇒ 2d² = 98
⇒ d² = 49 ⇒ d = ± 7
When a = 11 and d = 7 then required three numbers are 11 – 7, 11, 11 + 7 i.e. 4, 11, 18 When a = 11 and d = -7 then required numbers are 11 +7, 11, 11 – 7 i.e. 18, 11, 4.

Question 6.
There are four numbers in A.P., the sum of the two extremes is 8 , and the product of the middle two is 15 . What are the numbers?
Solution:
Let the four numbers in A.P are
a – 3d, a – d, a + d, a + 3d
according to given conditions, we have
a – 3d + a + 3d = 8 ⇒ 2a = 8 ⇒ a = 4
Also, (a – d)(a + d) = 15 ⇒ a² – d² = 15
⇒ 4² – d² = 15 ⇒ d² = 16 – 15 = 1
⇒ d = ± 1
When a = 4 and d = 1
Then the numbers are ; 4 – 3; 4 – 1; 4 + 1; 4 + 3 i.e. 1, 3, 5 and 7 .
When a = 4 and d = -1
Then required numbers are ; 4 + 3, 4 + 1, 4 – 1, 4 – 3 i.e. 7, 5, 3 and 1 .

Question 7.
The sum of the first three terms of an A.P. is 36 while their product is 1620 . Find the A.P.
Solution:
Let the three numbers in A.P are
a – d, a, a + d.
Since the sum of these numbers be 36 .
∴ a – d + a + a + d = 36 ⇒ 3a = 36
⇒ a = 12
and their product is 1620 .
(a – d) a (a + d) = 1620
⇒ 12(12² – d²) = 1620
⇒ 144 – d² = 135 ⇒ d² = 9 ⇒ d = ± 3
When a = 12 and d = 3
Then required numbers are;
12 – 3, 12, 12 + 3 i.e. 9, 12, 13
When a = 12 and d = – 3
Then required numbers are ;
12 + 3, 12, 12 – 3 i.e. 15, 12 and 9 .

Question 8.
The angles of a triangle are in A.P. If the greatest angle is double the least, find the angles.
Solution:
Let the angles of triangle are in A.P are
a – d° + a°, a + d°
Since the sum of angles of △ be 180°.
∴ a – d° + a° + a + d° = 180°
⇒ 3a° = 180° ⇒ a° = 60°
according to given condition, we have
a + d° = 2 (a – d°) ⇒ a = 3d° ⇒ d° = 20°
Thus, the required angles are 60° – 20°, 60°, 60° + 20° i.e. 40°, 60° and 80°.

Question 9.
The sum of the first three consecutive terms of an A.P. is 9 and the sum of their squares is 35. Find S_n.
Solution:
Let the three consecutive terms of an A.P are (a – d)°, a° and (a + d)°
Since their sum of 9 .
∴ (a – d)° + a° + (a + d)° = 9
⇒ 3a = 9 ⇒ a = 3
Also sum of their squares is 35 .
∴ (a – d)² + a² + (a + d)² = 35
⇒ 3a² + 2d² = 35
⇒ 27 + 2d² = 35
⇒ 2d² = 8
⇒ d² = 4
⇒ d = ± 2
When a = 3, d = 2, required numbers are
3 – 2, 3, 3 + 2 i.e. 1, 3, 5
When a = 3, d = -2, required numbers are
3 + 2, 3, 3 – 2 i.e. 5, 3, 1.

Question 10.
a₁, a₂, a₃, a₄, a₅ are first five terms of an A.P. such that a₁ + a₃ + a₅ = – 12 and a₁ a₂ a₃ = 8. Find the first term and the common difference.
Solution:
Let a₁ be the first term and d be the common difference of given A.P.
We know that
a_n = a₁ + (n – 1) d
given a₁ + a₃ + a₅ = – 12
⇒ a₁ + a₁ + 2d + a₁ + 4d = – 12
⇒ 3a₁ + 6d = – 12
⇒ a₁ + 2d = – 4
Also, a₁ a₂ a₃ = 8
⇒ a₁(a₁ + d) (a₁ + 2d) = 8 …(2)
From eqn. (1) and eqn. (2); we have
(- 4 – 2d)(- 4 – d)(- 4) = 8
⇒ (4 + 2d)(4 + d ) = – 2
⇒ (2 + d)(4 + d) = – 1
⇒ d² + 6d + 9 = 0
⇒ (d + 3)² = 0 ⇒ d = – 3
∴ from (1); a₁ – 6 = – 4 ⇒ a₁ = 2
Hence the required first term of given A.P. be 2 and common difference be -3 .

Question 11.
The angles of a quadrilateral are in A.P. and the greatest angle is double the first angle. Find the circular measure of the least angle.
Solution:
Let the angles of quadrilateral in A.P are (a – 3d)°, (a – d)°, (a + d)° and (a + 3d)°
Since the sum of all angles of quadrilateral be 360°.
∴ (a – 3d)° + (a – d)° + (a + d)° + (a + 3d)° = 360°
⇒ 4a = 360
⇒ a = 90
Also according to given condition, we have
(a + 3d)° = 2 (a – 3d)°
⇒ a = 9d
⇒ d = \(\frac { 90 }{ 9 }\) = 10
∴ least angle of quadrilateral be (a – 3d)° i.e. (90 – 3 × 10)° i.e. 60° i.e. \(\frac{\pi}{3}\) radians
[∵ π radians = 180°]

Question 12.
If a, b, c are in A.P., show that
a\(\left(\frac{1}{b}+\frac{1}{c}\right)\), b\(\left(\frac{1}{c}+\frac{1}{a}\right)\), c\(\left(\frac{1}{a}+\frac{1}{b}\right)\) are in A.P.
Solution:
Since a, b, c are in A.P
Dividing each term by abc
∴ \(\frac{a}{a b c}\), \(\frac{b}{a b c}\), \(\frac{c}{a b c}\) are in A.P
i.e. \(\frac{1}{bc}\), \(\frac{1}{ca}\), \(\frac{1}{ab}\) are in A.P.
Multiplying each term by ab + bc + ca, we have
\(\frac{a b+b c+c a}{b c}\), \(\frac{a b+b c+c a}{c a}\), \(\frac{a b+b c+c a}{a b}\) are in A.P.
⇒ \(\frac{a(b+c)}{b c}\) + 1, \(\frac{b(a+c)}{c a}\) + 1, \(\frac{c(a+b)}{a b}\) + 1 are in A.P.
Subtracting 1 from each term, we have
a\(\left(\frac{1}{c}+\frac{1}{b}\right)\), b\(\left(\frac{1}{c}+\frac{1}{a}\right)\), c\(\left(\frac{1}{b}+\frac{1}{a}\right)\) are in A.P.

Question 13.
If \(\frac { 1 }{ x }\), \(\frac { 1 }{ y }\), \(\frac { 1 }{ z }\) are A.P., show that
(i) yz, zx, xy are in A.P.
(ii) xy, zx, yz are in A.P.
(iii) \(\frac{y+z}{x}\), \(\frac{z+x}{y}\), \(\frac{x+y}{z}\) are in A.P.
Solution:
(i) Given \(\frac { 1 }{ x }\), \(\frac { 1 }{ y }\), \(\frac { 1 }{ z }\) are in A.P.
∴ \(\frac { 2 }{ y }\) = \(\frac { 1 }{ x }\) + \(\frac { 1 }{ z }\) ⇒ \(\frac { 2 }{ y }\) = \(\frac{z+x}{z x}\)
⇒ 2zx = yz + xy …(1)
∴ yz, xz, xy are in A.P
[∵ a, b, c are in A.P iff 2b = a + c]
(ii) Also from (1); xy, xz, yz are in A.P.
(iii) Since \(\frac { 1 }{ x }\), \(\frac { 1 }{ y }\), \(\frac { 1 }{ z }\) are in A.P.
Multiplying each term by x + y + z; we have
\(\frac{x+y+z}{x}\), \(\frac{x+y+z}{y}\), \(\frac{x+y+z}{z}\) are in A.P.
⇒ \(\frac{y+z}{x}\) + 1, \(\frac{x+z}{y}\) + 1, \(\frac{x+y}{z}\) + 1 are in A.P.
On subtracting 1 from each term
∴ \(\frac{y+z}{x}\), \(\frac{x+z}{y}\), \(\frac{x+y}{z}\) are in A.P.

Question 14.
If (b + c)^-1, (c + a)^-1, (a + b)^-1 are in A.P., then show that \(\frac{a}{b+c}\), \(\frac{b}{c+a}\), \(\frac{c}{a+b}\) are also in A.P.
Solution:
Let \(\frac{a}{b+c}\), \(\frac{b}{c+a}\), \(\frac{c}{a+b}\) are in A.P.
adding 1 to each term, we have
\(\frac{a}{b+c}\) + 1, \(\frac{b}{c+a}\) + 1, \(\frac{c}{a+b}\) + 1 are in A.P.
⇒ \(\frac{a+b+c}{b+c}\), \(\frac{b+c+a}{c+a}\), \(\frac{c+a+b}{a+b}\) are in A.P.
Dividing each term by a + b + c ≠ 0
⇒ \(\frac{1}{b+c}\), \(\frac{1}{c+a}\), \(\frac{1}{a+b}\) are in A.P.
which is true by given result.
Hence \(\frac{a}{b+c}\), \(\frac{b}{c+a}\), \(\frac{c}{a+b}\) are in A.P.

Question 15.
If \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}\), \(\frac{a+b-c}{c}\) are in A.P., then prove that \(\frac { 1 }{ a }\), \(\frac { 1 }{ b }\), \(\frac { 1 }{ c }\) are also in A.P.
Solution:
Given \(\frac{b+c-a}{a}\), \(\frac{c+a-b}{b}\), \(\frac{a+b-c}{c}\) are in A.P.
adding 2 to each term, we have
\(\frac{b+c-a}{a}\) + 2, \(\frac{c+a-b}{b}\) + 2, \(\frac{a+b-c}{c}\) + 2 are in A.P.
⇒ \(\frac{b+c+a}{a}\), \(\frac{b+c+a}{b}\), \(\frac{a+b+c}{c}\) are in A.P.
dividing each term by (a + b + c); we have
\(\frac { 1 }{ a }\), \(\frac { 1 }{ b }\), \(\frac { 1 }{ c }\) are in A.P.

Question 16.
If x, y, z are in A.P., show that (x y)^-1,(zx)^-1,(yz)^-1 are also in A.P.
Solution:
Given x, y, z are in A.P.
⇒ z, y, x are in A.P.
dividing each term by xyz, we have

Question 17.
If a², b², c² are in A.P., prove that \(\frac{a}{b+c}\), \(\frac{b}{c+a}\), \(\frac{c}{a+b}\) are in A.P.
Solution: