## 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

∴ T_{6} = 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

∴ T_{18} = a + 17d = 9 + 17 × 10

= 9 + 170 = 179

Hence, 18th term of corresponding H.P. is = \(\frac { 1 }{ 179 } \)

Hence, 10^{th} term of corresponding H.P. is = \(\frac { 2 }{ 37 } \)

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 = \(\frac { 2 }{ 11-n } \)

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 = \(\frac { 2 }{ na+(2-n)b } \)

Question 3.

Find that H.P. whose 2nd term is \(\frac { 2 }{ 5 } \) and 7th term is \(\frac { 4 }{ 25 } \)

Solution:

Here 2^{nd} term of corresponding A.P. is \(\frac { 5 }{ 2 } \) and 7th term is \(\frac { 25 }{ 4 } \)

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 \(\frac { 2 }{ 17 } \) and \(\frac { 2 }{ 13 } \)

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 H_{1}, H_{2}, H_{3}, H_{4} are 4 H.M. between 1 and \(\frac { 1 }{ 16 } \).

So, 1, H_{1}, H_{2}, H_{3}, H_{4}, \(\frac { 1 }{ 16 } \) are in H.P.

First term of corresponding A.P. is 1 and 6^{th} term is 16.

∴ a + 5d = 16 (∵ T_{6} = 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 ax^{2} + bx + c = 0.

Solution:

ax^{2} + bx + c = 0

Let its root are α and β

Hence, H.M. of roots of quadratic equation will be \(\frac { -2c }{ b } \)

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 \(\frac { pq }{ (p+q) } \)

Solution:

Let p^{th} term of corresponding A.P. is \(\frac { 1 }{ q } \) and q^{th} term is \(\frac { 1 }{ q } \) Let first term is a and common difference is d.

Question 11.

If roots of equation

a(b – c)x^{2} + 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 B^{2} – 2AC = 0

i.e., [b(c – a)^{2} – 4a(b – c) × c(a – b) = 0

or b^{2} (c^{2} + a^{2} – 2ca) – 4ac (ah – b^{2} – ca + bc) = 0

or b^{2} (c^{2} + a^{2} – 2ca + 4ca) – 4abc(a + c) + 4a^{2} c^{2} = 0

or b^{2}_{ }(c^{2} + a^{2} + 2ca) – 4abc (a + c) + (2ac)^{2} = 0

or b^{2} (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:

## Leave a Reply