RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7

Question 1.
Find the term mentioned in following Harmonic Progressin –
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
Solution:
(i) The A.P. corresponding to given H.P. is
2, 5, 8, 11, …
For this A.P. a = 2. d = 5 – 2 = 3
∴ T6 = a + 5d = 2 + 5 × 3
= 2 + 15 = 17
hence, 6th term of corresponding H.P. is = 1/17.

(ii) The A.P. corresponding to given H.P.
9, 19, 29, 39, …
For AP. a = 9, d = 19 – 9 = 10
∴ T18 = a + 17d = 9 + 17 × 10
= 9 + 170 = 179
Hence, 18th term of corresponding H.P. is = \(\frac { 1 }{ 179 } \)
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
Hence, 10th term of corresponding H.P. is = \(\frac { 2 }{ 37 } \)

Question 2.
Find nth term of the following H.P.:
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
Solution:
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
The A.P. corresponding to given H.P. is
10, 9, 8, 7, …
For this A.P. a = 10, d = 9 – 10 = – 1
∴ Tn = a + (n – 1)d
= 10 + (n – 1)(- 1)
= 10 – n + 1 = 11 – n
Hence, nth term of corresponding H.P. is = \(\frac { 2 }{ 11-n } \)
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
The A.P. corresponding to given H.P. is
a + b, 2a, 3a – b, …
For this A.P. a = a + b, d = 2a – a – b = a – b
∴ Tn = a + (n – 1)d
= (a + b) + (n – 1) (a – b)
= a + b + na – a – nb + b
= na + 2b – nb
= na + (2 – n)b
Hence, nth term of corresponding H.P. is = \(\frac { 2 }{ na+(2-n)b } \)

Question 3.
Find that H.P. whose 2nd term is \(\frac { 2 }{ 5 } \) and 7th term is \(\frac { 4 }{ 25 } \)
Solution:
Here 2nd term of corresponding A.P. is \(\frac { 5 }{ 2 } \) and 7th term is \(\frac { 25 }{ 4 } \)
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7

Question 4.
If 7th term of H.P. is 17/2 and 11th term is 13/2 then find 20th term.
Solution:
The 7th and 9th term of corresponding A.P. will be \(\frac { 2 }{ 17 } \) and \(\frac { 2 }{ 13 } \)
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
Hence, 20th term of corresponding H.P. will be \(\frac { 17 }{ 4 } \)

Question 5.
Find :
(i) 4 H.M. between 1 and 1/16.
(ii) 5 H.M. between 1/19 and 1/7.
(iii) 1 H.M. between – 2/5 and 4/25.
Solution:
(i) Let H1, H2, H3, H4 are 4 H.M. between 1 and \(\frac { 1 }{ 16 } \).
So, 1, H1, H2, H3, H4, \(\frac { 1 }{ 16 } \) are in H.P.
First term of corresponding A.P. is 1 and 6th term is 16.
∴ a + 5d = 16 (∵ T6 = a + 5d)
⇒ 1 + 5d = 16
⇒ 5d = 15 ⇒ d = 3 (∵ a2)
So. there will be 4 AM. between 1 and 16.
1 + d, 1 + 2d, 1 + 3d, 1 + 4d
or 1 + 3, 1 + 2(3), 1 + 3(3), 1 + 4(3)
or 4, 7, 10, 13
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
So, corresponding A.M. are a + d, a + 2d, a + 3d, a + 4d, a + 5d
or 19 – 2, 19 + 2(- 2), 19 + 3(- 2), 19 + 4(- 2), 19 + 5(- 2)
or 17, 15, 13, 11,9
Hence, corresponding H.M. will be
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7

Question 6.
If a, b, c are pth, qth and rth terms of H.P. respectively then prove that
bc(q – r) + ca(r – p) + ab(p – q) = 0.
Solution:
The pth term of H.P. = a
ath term = b
rth term = c
Let first term of corresponding A.P. is x and common difference is y. Then
pth term of corresponding
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7

Question 7.
If a, b, c are in H.P., then prove that a, a – c, a – b will be in H.P.
Solution:
Let x = a. y = a – c, z = a – b are in HP. Then by the property of H.P.
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
Thus, a, b, c are in H.P. which are given. Hence, our hypothesis i.e., a, a – c, a – b are in H.P. is true.

Question 8.
It a, b, c are in H.P., then prove that
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
Solution:
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7

Question 9.
Find H.M. of roots of quadratic equation ax2 + bx + c = 0.
Solution:
ax2 + bx + c = 0
Let its root are α and β
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
Hence, H.M. of roots of quadratic equation will be \(\frac { -2c }{ b } \)

Question 10.
If pth term of any H,P. is q and qth term is p, then prove that (p + q) th term will be \(\frac { pq }{ (p+q) } \)
Solution:
Let pth term of corresponding A.P. is \(\frac { 1 }{ q } \) and qth term is \(\frac { 1 }{ q } \) Let first term is a and common difference is d.

RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7

Question 11.
If roots of equation
a(b – c)x2 + b(c – a)x + c(a – b) = 0
are same, then prove that a, b, c will be in H.P.
Solution:
Roots of given equation are same
so, Discriment B2 – 2AC = 0
i.e., [b(c – a)2 – 4a(b – c) × c(a – b) = 0
or b2 (c2 + a2 – 2ca) – 4ac (ah – b2 – ca + bc) = 0
or b2 (c2 + a2 – 2ca + 4ca) – 4abc(a + c) + 4a2 c2 = 0
or b2 (c2 + a2 + 2ca) – 4abc (a + c) + (2ac)2 = 0
or b2 (a + c)2 – 2(2ac) [b(a + c)] + (2ac)2 = 0
⇒ [b(a + c) – 2ac]2 = 0
or b(a+ c) – 2ac = 0
or b = \(\frac { 2ac }{ a+c } \)
Hence, a, b, c are in H.P.

Question 12.
If a student, from his house goes to school at the speed of 8 km/h and returns with the speed of 6 km/h, then find its average speed where as distance between house and school is 6 km. Also verify the answer.
Solution:
RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.7

RBSE Solutions for Class 11 Maths