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 –

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, 6^th 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 =

Hence, 10^th term of corresponding H.P. is =

Question 2.
Find n^th term of the following H.P.:

Solution:

The A.P. corresponding to given H.P. is
10, 9, 8, 7, …
For this A.P. a = 10, d = 9 – 10 = – 1
∴ T_n = a + (n – 1)d
= 10 + (n – 1)(- 1)
= 10 – n + 1 = 11 – n
Hence, n^th term of corresponding H.P. is =

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
∴ T_n = a + (n – 1)d
= (a + b) + (n – 1) (a – b)
= a + b + na – a – nb + b
= na + 2b – nb
= na + (2 – n)b
Hence, n^th term of corresponding H.P. is =

Question 3.
Find that H.P. whose 2nd term is and 7th term is
Solution:
Here 2^nd term of corresponding A.P. is and 7th term is

Question 4.
If 7^th term of H.P. is 17/2 and 11^th term is 13/2 then find 20th term.
Solution:
The 7^th and 9^th term of corresponding A.P. will be and


Hence, 20th term of corresponding H.P. will be

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 .
So, 1, H₁, H₂, H₃, H₄, are in H.P.
First term of corresponding A.P. is 1 and 6^th 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


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

Question 6.
If a, b, c are p^th, q^th and r^th terms of H.P. respectively then prove that
bc(q – r) + ca(r – p) + ab(p – q) = 0.
Solution:
The p^th term of H.P. = a
a^th term = b
r^th term = c
Let first term of corresponding A.P. is x and common difference is y. Then
p^th term of corresponding

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.

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

Solution:

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

Hence, H.M. of roots of quadratic equation will be

Question 10.
If p^th term of any H,P. is q and q^th term is p, then prove that (p + q) th term will be
Solution:
Let p^th term of corresponding A.P. is and q^th term is Let first term is a and common difference is d.

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 =
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