RBSE Maths Class 11 Chapter 9: Important Questions and Solutions

RBSE Solutions For Class 11 Maths Chapter 9 – Sequence, progression and series available here. The important questions and solutions of Chapter 9, available at BYJU’S, contain step by step explanations for the better understanding of the students. All these important questions are based on the new pattern prescribed by the RBSE. Students can also get the syllabus and textbooks on RBSE Class 11 solutions.

Chapter 9 of the RBSE Class 11 Maths will help the students to solve problems related to the sequence, series, progression, arithmetic progression, sum to n terms of an arithmetic progression, arithmetic mean, properties of arithmetic progression, geometric progression, geometric mean, the relation between arithmetic and geometric mean, sum to n terms of a geometric progression, the sum of an infinite GP, arithmetic and geometric series, sum to n terms of a series, harmonic progression, harmonic mean, relation between the means. After referring to these important questions and solutions, you can refer to 2020-2021 solutions for the chapter here.

RBSE Maths Chapter 9: Exercise 9.1 Textbook Important Questions and Solutions

Question 1: From the following sequences which are in A.P.?

(i) 2, 6, 11, 17,…

(ii) 1, 1.4, 1.8, 2.2,…

(iii) -7, -5, -3, -1,..

(iv) 1, 8, 27, 64,…

Solution :

Given sequence will be in A.P. only if the common difference of two consecutive terms is the same.

Question 2: Find the first term, common difference and the 5^th term of given sequences which have the following n^th terms:

(i) 3n+ 7

(ii) a + (n – 1) d

(iii) 5 – 3n

Solution:

(i) 3n + 7

n^th term, T_n = 3n + 7

First term T₁ = 3 × 1 + 7 = 3 + 7 = 10

Second term T₂ = 3 × 2 + 7 = 6 + 7 = 13

Common difference

d = T₂ – T₁

= 13 – 10

= 3

Fifth term T₅ = 3 × 5 + 7 = 15 + 7 = 22

Hence, a = 10, d = 3 and T₅ = 22

(ii) a + (n – 1) d

n^th term T_n = a + (n – 1) d

First term T₁ = a + (1 – 1) d

= a + 0 × d

= a + 0

= a

Second term T₂ = a + (2 – 1) d

= a + d

Common difference

d = T₂ – T₁

= a + d – a

= d

Fifth term T₅ = a + (5 – 1) d

= a + 4d

Hence, T₅ = a + 4d

(iii) 5 – 3n

n^th term T_n = 5 – 3n

First term,

T₁ = 5 – 3 × 1

= 5 – 3

= 2

Second term

T₂ = 5 – 3 × 2

= 5- 6

= -1

Common difference

d = T₂ – T₁

= (- 1) – 2

= -3

Fifth term

T₅ = 5 – 3 × 5

= 5 – 15

= – 10

Hence, a = 2, d = -3 and T₅ = – 10.

Question 3: Show that the following sequences with their n^th terms are not in arithmetic progression.

[i] [n] / [n + 1]

[ii] n² + 1

Solution:

If the value of T_{n +1} – T_n is not independent of n then the sequence of n^th terms are not in arithmetic progression.

T_{n +1} – T_n = {[n + 1] / [n + 1] + 1} – {[n] / [n + 1]}

= {[n + 1] / [n + 2]} – {[n] / [n + 1]}

= {[n + 1]² – [n(n +2)]} / [(n + 1) (n + 2)]

= {(n² + 2n + 1 – n² – 2n)} / [(n + 1) (n + 2)]

= 1 / [(n + 1) (n + 2)]

Hence, the given sequence with its n^th term is not in arithmetic progression.

T_{n +1} – T_n = {(n+ 1)² + 1} – {n² + 1}

= (n² + 1 + 2n + 1) – (n² + 1)

= (n² + 2n + 2 – n² – 1)

= 2n + 1

Hence, the given sequence with its n^th term is not in arithmetic progression.

Question 4: In an A.P. 2 + 5 + 8 + 11 +… which term is 65?

Solution:

2 + 5 + 8 + 11 + ….

Let the n^th term be 65.

Then T_n = 65 and d = 5 – 2 = 3

T_n = a + (n – 1) d

⇒ a + (n – 1) d = 65

⇒ 2 + (n – 1) × 3 = 65

⇒ 2 + 3n – 3 = 65

⇒ 3n – 1 = 65

⇒ 3n = 65 + 1

⇒ 3n = 66

⇒ n = 66 / 3

⇒ n = 22

Hence, the 22^nd term is 65.

Question 5: In an A.P. 4 + 9 + 14 + 19 +… + 124, find the 13^th term from the last.

Solution:

4 + 9 + 14 + 19 +…+ 124

a = 4, d = 9 – 4 = 5, L = 124

13^th term from last

= L – (n – 1) d

= 124 – (13 – 1) × 5

= 124 – 12 × 5

= 124 – 60

= 64

Hence, the 13^th term from the last is 64.

Question 6: In an A.P. 2 + 5 + 8 + 11 + … if the last term is 95, then find the number of terms of the series.

Solution:

2 + 5 + 8 + 11 + … + 95

a = 2, d = 5 – 2 = 3, l = 95

Then, from l = a + (n – 1 ) d

⇒ 2 + (n – 1) × 3 = 95

⇒ 3 (n – 1) = 95 – 2

⇒ (n – 1) = 93 / 3

⇒ (n – 1) = 31

⇒ n = 31 + 1

= 32

Hence, the number of terms of the series is 32.

Question 7: If the 9^th term of A.P. is zero, then prove that 29^th term is twice the 19th term.

Solution:

Let a be the first term and d be the common difference then 9^th term = T₉ = 0

Then a + (9 – 1) d= 0

⇒ a + 8d = 0 ….. (i)

29^th term = T₂₉ = a + 28d … (ii)

19^th term = T₁₉ = a + 18d … (iii)

Putting the value of equation (i) into equation (ii) and (iii)

T₂₉ = (a + 8d) + 20d

⇒ T₂₉ = 0 + 20 d

⇒ T₂₉ = 20d … (iv)

⇒ T₁₉= (a + 8 d) + 10 d

⇒ T₁₉ = 0 + 10d

⇒ T₁₉ = 10d … (v)

From equation (iv) and (v), we have

T₂₉ = 20d = 2 × 10d = 2 × T₁₉

T₂₉ = 2 T₁₉

Question 8: How many two-digit natural numbers are divisible by 3?

Solution:

The first two-digit number which is divisible by 3 is 12 and the last two-digit number is 99.

To find the numbers which are divisible by 3, the common difference is 3.

a = 12, d = 3, l = 99

Hence, l = a + (n – 1 )d

⇒ 12 + (n – 1) × 3 = 99

⇒ 3 (n – 1) = 99 – 12

⇒ n = 29 + 1

= 30

Hence, the two digit natural numbers which are divisible by 3 are 30.

Question 9: If in A.P., p^th term is q and q^th term is p, then find (p + q)^th term.

Solution:

Let the first term of A.P. = a and common difference = d

Then p^th term = T_p = a + (p – 1) d = q

⇒ a + (p – 1) d = q … (i) and

q^th term = T_p = a + (q – 1) d = p

⇒ a + (q – 1)d = p …. (ii)

(p^th + q^th term) = T_p+q = a+ {(p + q) -1} …. (iii)

Subtracting equation (ii) from (i),

{a + (p – 1)d} – {a + (q – 1)d} = q – p

⇒ a + (p – 1) d – a – (q – 1)d = q – p

⇒ (p – 1 – q + 1 )d = q – p

⇒ (p – q)d – (p – q)

⇒ d = -1

Putting value of d in equation (i)

a + (p- 1) (- 1) = q

⇒ a – (p – 1) = q

⇒ a = q + p – 1

⇒ a = p + q – 1

Putting the value of a and d in equation (iii),

T_p+q= a + [(p + q) -]d

= (p + q – 1) + [(p + q -1)(- 1)]

= (p + q -1) – (p + q – 1)

= 0

Hence, (p + q)^th term is zero.

Question 10: In an A.P. p^th term is 1/q and 17^th term 1/p, then prove that pq^th term is unity.

Solution:

Let the first term be a and common difference be d.

p^th term = T_p = a + (p – 1) d = 1 / q

a + (p – 1) d = 1 / q —- (1)

q^th term = T_q = a + (q – 1) d = 1 / p

a + (q – 1) d = 1 / p —- (2)

Solving equations (1) and (2),

d = 1 / pq and a = 1 / pq

pq^th term T_pq = a + (pq – 1) d —- (3)

Subtract equation (2) from (1),

a + (p – 1) d = 1 / q —- (1)

a + (q – 1) d = 1 / p —- (2)

(pd – d) – (qd – d) = [1 / q] – [1 / p]

(p – q)d = [p – q] / pq

d = 1 / pq

Putting the value of d in equation (1),

a + (p – 1) * (1 / pq) = 1 / q

a + (1 / q) – (1 / pq) = 1 / q

a = 1 / pq

Putting the values of a and d in equation (3),

a_pq = a + (pq – 1) d

a_pq = [1 / pq] + (pq – 1) * (1 / pq)

a_pq = 1

RBSE Maths Chapter 9: Exercise 9.2 Textbook Important Questions and Solutions

Question 1: Find the sum of the following progression 7 + 11 + 15 + 19 + …… upto 20 terms.

Solution:

7 + 11 + 15 + 19 + …

Here, a = 7, d = 11 – 7 = 4, n = 20

Then S_n = [n / 2] * [2a + (n – 1)d]

S₂₀ = [20 / 2] * [2 * 7 + (20 – 1) * 4]

= 10 * [14 + 19 × 4]

= 10 × (14 + 76)

= 10 × 90

= 900

Hence, the sum of 20 terms is 900.

Question 2: Find the sum of the odd integers from 1 to 101, which is divisible by 3.

Solution:

Odd integers from 1 to 101 are 1, 3, 5, 7, 9, …, 101

Integers which are divisible by 3 are 3, 9, 15, 21, …… 99.

First term of this series a = 3, common difference d = 9 – 3 = 6, last term = 99 and let the number at the last term be n, then the last term

= 3 + (n – 1) × 6 = 99 [From formula, l = a + (n + 1 )d]

⇒ 6 (n – 1) = 99 – 3

⇒ (n – 1) = 96 / 6

⇒ n = 16 + 1

= 17

Hence, the sum of 17 terms

S_n = [n / 2] * [2a + (n – 1)d]

S₁₇ = [17 / 2] * [2 * 3 + (17 – 1) * (6)]

= [8.5] * [6 + 96]

= 8.5 * 102

= 867

Hence, the sum of odd integers from 1 to 101, which is divisible by 3 is 867.

Question 3: Find the sum of n terms of A.P. whose r^th term is 2r + 3.

Solution:

T_r = 2r + 3

Then, T_n = (2n + 3)

Hence, T₁ = 2 × 1 + 3

= 2 + 3

= 5 and

T₂ = 2 × 2 + 3

= 4 + 3

= 7

Common difference

d = T₂ – T₁

= 7 – 5

= 2

∴ Sum of n terms, S_n = [n / 2] * [2a + (n – 1)d]

= [n / 2] * [2 * 5 + (n – 1) * 2]

= [n / 2] * [10 + 2n – 2]

= [n / 2] * [8 + 2n]

= n [4 + n]

Hence, the sum of n terms is n (n + 4).

Question 4: If the sum of n terms of A.P. 1, 6, 11,…. is 148, then find the number of terms and last term.

Solution:

Given progression = 1, 6, 11, …

S_n = 148, a = 1,d = 6 – 1 = 5

∵ S_n = [n / 2] * [2a + (n – 1)d]

⇒ [n / 2] * [2 × 1 + (n – 1) × 5] = 148

⇒ [n / 2] * [2 + 5n – 5] = 148

⇒ n (5n – 3) = 296

⇒ 5n² – 3n – 296 = 0

On solving,

⇒ 5n² – 40n + 37n – 296 = 0

⇒ 5n( n – 8) + 37( n – 8) = 0

⇒ (n – 8) (5n + 37) = 0

⇒ n – 8 = 0 or 5n + 37 = 0

⇒ n = 8 or n = -37 / 5

Since the number of terms cannot be fraction or a negative number, n = 8.

Therefore n is the 8^th term.

T_n = T₈ = a + (n – 1) d

= 1 + (8 – 1) 5

= 1 + 7 (5)

= 1 + 35

= 36

8^th term = n = 36.

Question 5: If the sum of n, 2n, 3n terms of any A.P. are S₁, S₂ and S₃ respectively, then prove that S₃ = 3 (S₂ – S₁).

Solution:

Let a be the first term and d be the common difference of an A.P. then,

Sum to n terms = S₁ = [n / 2] * [2a + (n – 1)d] —– (1)

Sum to 2n terms = S₂ = [2n / 2] * [2a + (2n – 1)d] —- (2)

Sum to 3n terms = S₃ = [3n / 2] * [2a + (3n – 1)d] —- (3)

Subtracting equation (1) from equation (3),

Multiplying by 3 on both sides,

3 (S₂ – S₁) = [3n / 2] [2a + (3n – 1) d]

= S₃ [from equation (3)]

Therefore, S₃ = 3 (S₂ – S₁).

Question 6: Find the sum of A.P. a₁ + a₂ + a₃…, a₃₀, given that a₁ + a₇ + a₁₀ + a₂₁ + a₂₄ + a₃₀ = 540.

Solution:

The number of terms in series = 30

T_r +T_{r – 1} = a + 1

∵ a₇ is 7^th term from starting and a₂₄ is 7^th term from the end,

a₇ + a₂₄ = a₁+ a₃₀ …(i)

Similarly, a₁₀ is 10^th term from starting and a₂₁ is 10th term from the end, then

a₁₀ + a₂₁ = a₁+ a₃₀

a₁ + a₇ + a₁₀ + a₂₁ + a₂₄ + a₃₀ = 540

⇒ (a₁ + a₃₀) + (a₇ + a₂₄) + (a₁₀+ a₂₁) = 540

From equation (i) and (ii),

(a₁ + a₃₀) + (a₁ + a₃₀) + (a₁+ a₃₀) = 540

⇒ (a₁ + a₃₀) = 540

⇒ (a₁ + a₃₀) = 180

Hence, sum of 30 terms,

S₃₀ = (a₁ + a₃₀)

= 15 (180)

= 2700

Hence, a₁ + a₂ + a₃…, a₃₀ = 2700.

Question 7: The interior angles of a polygon are in A.P. The smallest interior angle is 52° and if the difference of consecutive interior angles is 8°, then find the number of sides of the polygon.

Solution:

Smallest angle = 52°

The difference between consecutive angles = 8°

Let the number of sides of a polygon be x.

First-term a = 52°

Common difference, d = 8°

Sum of interior angles = (n – 2) * 360°

S₁ = [n / 2] * [2a + (n – 1)d]

⇒ (n – 2) 360° = [n / 2] * [2 x 52° + (n – 1) 8°]

⇒ 360°n – 720° = 104°n + 8°n² – 8°n

⇒ 8°n² + (104° – 360° – 8°)n + 720° = 0

⇒ 8°n² – 264°n + 720° = 0

⇒ n² – 33°n + 90° = 0

On solving, n = 3 or 30

But n ≠ 30 because, for n = 30, we get

Last angle d_n = a + (n + 1)d

= 52° + (30 – 1)8°

= 52° + 29 × 8° = 52° + 232°

= 284° which is impossible

Hence, the number of sides is 3.

RBSE Maths Chapter 9: Exercise 9.3 Textbook Important Questions and Solutions

Question 1: Find the 7^th term of the series 1 + 3 + 9 + 27 + ……..

Solution:

Given series = 1 + 3 + 9 + 27 + ….

Here, a = 1, r = 3 / 1 = 3

7^th term = T₁ (From T_n = ar^n-1)

= 1 × (3)^7-1

= (3)⁶

= 729

Question 2: Which term of series 64 + 32 + 16 + 8 + … is 1 / 64?

Solution:

64 + 32 + 16 + 8 + …

Let the n^th term be 1 / 64.

a = 64, r = 32 / 64 = [1 / 2]

T_n = ar^n-1

On comparing the powers, since the bases are the same, we get,

n – 1 = 12

n = 12 + 1

n = 13

The 13^th term is [1 / 64].

Question 3: The third term of a G.P. is 32 and 7^th term is 8192, then find 10^th term of GP.

Solution:

T₃ = ar² = 32 —- (1)

T₇ = ar⁶ = 8192 —– (2)

Divide equation (ii) by (i), we get

⇒ r⁴ = 256

⇒ r⁴ = 4⁴

On comparing r = 4

Then, from equation (i)

ar² = 32

⇒ a × 4² = 32

⇒ a = 2

10^th term

= T₁₀

= ar⁹

= 2 × (4)⁹

= 524288

Question 4: Find the three G.M. between 3 and 48.

Solution:

Let G₁, G₂, G₃ be the three GM between 3 and 48, then

3, G₁, G₂, G₃, 48 will be in G.P.

a = 3, last term (5^th term) = 48 and n = 5

3 * r⁴ = 48 [∵ l = ar^{n – 1} ]

⇒ r⁴ = 16 = 2⁴

⇒ r = 2

G₁ = 3 × 2 = 6

G₂ = 3 × 22 = 12

G₃ = 3 × 23 = 24

Hence, the three G.M. are 6, 12, 24.

Question 5: If 4^th term of any G.P. is p, 7^th term is q and 10^th term is r, then prove that q² – pr.

Solution:

Let a is the first term and r is the common ratio of G.P., then according to the question

T₄ = ar^{4 – 1}

p = ar³ ….. (i)

T₇ = ar^{7 – 1}

q = ar⁶ ….. (ii)

T₁₀ = ar^{10 – 1}

r = ar⁹ ….. (iii)

Multiplying equation (i) and (iii), we get

a² r¹² = pr ….. (iv)

Squaring of equation (ii),

q² = a²r¹²

Hence, from equation (iv) and (v)

q² = pr

Question 6: If a, b, c are in G.P. and a^x = b^y = c^z, then prove that [1 / x] + [1 / z] = [2 / y].

Solution:

a^x = b^y = c^z = k

Then, a = k^1/x

b = k^1/y

c = k^1/z and a, b and c are in GP.

b² = ac —- (1)

Putting the value of a, b and c in equation (1),

k^2/y = k^1/x+1/z is possible when [1 / x] + [1 / z] = [2 / y].

Question 7: If n G.M. inserted between a and b then prove that the product of all geometric means will be (√xy)ⁿ.

Solution:

Let G₁, G₂, G₃, G₄, ……G_n are in G.M. in between x and y then

RBSE Maths Chapter 9: Exercise 9.4 Textbook Important Questions and Solutions

Question 1: Find the sum of the geometric progression 2 + 6 + 18 + 51 + …….. upto 7 terms.

Solution:

2 + 6 + 18 + 51 + …….. upto 7 terms

a = 2, r = 6 / 2 = 3, n = 7 {r > 1}

S_n = a[rⁿ – 1] / [r – 1]

S₇ = 2[3⁷ – 1] / [3 – 1]

S₇ = 2[3⁷ – 1] / 2

S₇ = [3⁷ – 1]

S₇ = 2187 – 1

= 2186

Question 2: Find the sum of the geometric progression 2 + 6 + 18 + 54 + … + 486.

Solution:

2 + 6 + 18 + 54….+ 486

a = 2, r = 6 / 2 = 3, T_n = 486 (r > 1)

n^th term T_n = ar^{n – 1}

ar^{n – 1} = 486

2 * (3)^{n – 1} = 486

⇒ (3)^{n – 1} = 243 = (3)⁵

⇒ n – 1 = 5

n = 5 + 1 = 6

Sum of n terms

S_n = a[rⁿ – 1] / [r – 1]

S₆ = 2 [3⁶ – 1] / [3 – 1]

S₆ = 3⁶ – 1

= 729 – 1

= 728

Hence, the sum is 728.

Question 3: Find the sum of first n terms of the following series 7 + 11 + 777 + …

Solution:

7+ 77 + 777 + 7777 upto n terms.

It is clear that this series is not in G.P. but it can be related to G.P. by writing it in the following form.

S_n = 7 + 77 + 777 + 7777 + … upto n terms

= 7[1 + 11 + 111 + 1111 + … upto n terms]

= [7 / 9] [9 + 99 + 999 + …. upto n terms]

= [7 / 9] [(10 – 1) + (10² – 1) + (10³ – 1) …… upto n terms]

= {70[10ⁿ – 1] / 81} – [7n / 9]

Question 4: Find the rational form of the recurring decimal 2.3555555.

Solution:

Digit 5 is repeated in 2.35.

∴ 2.35 = 2.35555…………

= 2 + [.3 + .05 + .005 + .0005+ … upto infinity]

Question 5: If y = x + x² + x³ + … ∞, where | x | < 1, then prove that x = [y] / [1 + y].

Solution:

y = x + x² + x³ + … ∞

⇒ y = x ( 1 + x + x² + … ∞)

⇒ y = x (1 / [1 – x])

⇒ y (1 – x) = x

⇒ y – xy – x =0

⇒ y – x(y + 1) =0

⇒ x(1 + y) = y

⇒ x = [y] / [1 + y]

RBSE Maths Chapter 9: Exercise 9.5 Textbook Important Questions and Solutions

Question 1: Find the sum of n terms of the series 1 + 1 + [3 / 2²] + [4 / 2³] + ……

Solution:

Let the sum of the series be S_n.

S_n = 1 + [2 / 2] + [3 / 2²] + [4 / 2³] + ……

In this, 1, 2, 3, 4 ….. is AP and 1, [1 / 2], [1 / 2²], ….. is a GP.

n^th term of an AP = 1 + (n – 1)n = n

n^th term of a GP = 1 * (1 / 2)^n-1 = 1 / 2^n-1

The arithmetic-geometric series is

Question 2: Find the sum of infinite terms of the series [3 / 7] + [5 / 21] + [7 / 63] + [9 / 189] ……

Solution:

Let the sum of series be S_∞.

The sum of the series is [6 / 7].

Question 3: Find n^th term and the sum of n terms of the series 3.2 + 5.2² + 7.2³ + …

Solution:

In the given series, A.P. is 3, 5, 7, …

The n^th term = 3 + (n – 1) 2

= 3 + 2n – 2

= 2n + 1

The G.P. is 2, 22 , 23 , … and its nth term is 2ⁿ.

Let the sum of its n terms be S_n and n^th term be T_n.

The n^th term of given arithmetic-geometric series is T_n = (2n + 1)2ⁿ.

Let the sum of series be S_n on putting the value of n in this, we get

S_n = 3.2 + 5.2² + 7.2³ + … + … + (2n – 1)2^{n + 1} + (2n+ 1)2ⁿ …(i)

2S_n = 3.2² + 5.2² + 7.2³ …+ (2n – 1)2ⁿ + (2n + 1)2^{n + 1} …(ii)

Subtracting equation (ii) from (i), we get

-Sn = 3.2 + 2[2² + 2³ + …2ⁿ] – (2n+ 1)^{2n + 1}

⇒ -Sn = 6 + 2² (2^{n + 1} – 1) – (2n + 1)^{2n + 1}

⇒ -Sn = 6 + 2^{n + 2} – 8 – (2n + 1)^{2n + 1}

⇒ -Sn = 2ⁿ⁺¹ [2 – 2n – 1] – 2

⇒ -Sn = (2n- 1)^{2n + 1} + 2

Hence, the sum of n terms = (2n + 5)^{2n + 1} + 2.

RBSE Maths Chapter 9: Exercise 9.6 Textbook Important Questions and Solutions

Question 1: Find the sum of n terms of the series whose n^th term is 3n² + 2n + 5.

Solution:

Question 2: Find the sum of the terms of the series 3² + 7² + 11² + 15² + …

Solution:

3² + 7² + 11² + 15² + …

n^th term of the given series T_n = [3 + (n – 1)4]²

= (3 + 4n – 4)²

= (4n – 1)²

= 16n² + 1 – 8n

∴ S_n = ∑(16n² – 8n + 1)

⇒ S_n = 16∑n² – 8∑M + ∑1

Question 3: Find the n^th term and sum of n terms of the series 1.3 + 3.5 + 5.7 + …

Solution:

1.3 + 3.5 + 5.7 + …

n^th term of the given series T_n = (1 + 3 + 5 + … n^th term)

(3 + 5 + 7 + … n^th term)

= [1 + (n – 1)² [3 +(n – 1)2]

= (1 + 2n – 2) (3 + 2n – 2)

= (2n – 1) (2n + 1)

∴ Sn = ∑ (4n² – 1)

= 4∑n² – ∑1

= 4 {[n (n + 1) (2n + 1)] / 6} – n

= [[2 / 3] n (n + 1) (2n + 1)] – n

= [n / 3] [2 (n + 1) (2n + 1) – 3]

= [n / 3] [4n² – 6n + 1 – 2]

= [n / 3] [4n² + 6n -1]

RBSE Maths Chapter 9: Exercise 9.7 Textbook Important Questions and Solutions

Question 1: Find the term mentioned in the harmonic progression [1 / 2], [1 / 5] , [1 / 8], [1 / 11] …. 6^th term.

Solution:

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, the 6^th term of corresponding H.P. is = 1 / 17.

Question 2: Find the n^th term of the HP [1 / 5], [2 / 9], [1 / 4], [2 / 7] ……

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, the n^th term of corresponding H.P. is = 2 / [11 – n].

Question 3: Find the 4 harmonic means between 1 and 1 / 16.

Solution:

Let H₁, H₂, H₃, H₄ are the 4 H.M. between 1 and 1 / 16.

So, 1, H₁, H₂, H₃, H₄ are in H.P.

The 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

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, the required HM is [1 / 4], [1 / 7], [1 / 10], [1 / 13].

Question 4: 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.

{[a – (a – c)] / [a – c] – [a – b]} = a / [a – b]

c / [b – c] = a / [a – b] [a – b] / [b – c] = a / c

Thus, a, b, c are in H.P. which are given.

Hence, by the hypothesis i.e., a, a – c, a – b is in H.P. is true.

Question 5: If a, b and c are in HP, then prove that [1 / (b – a)] + [1 / (b – c)] + [1 / a] + [1 / c] = [2 / b].

Solution:

RBSE Maths Chapter 9: Exercise 9.8 Textbook Important Questions and Solutions

Question 1: A.M. of two numbers is 60 and H.M. is 18, find the numbers.

Solution:

Let the two numbers are a and b, then

⇒ a + b = 100 and

From equation (i) and (ii)

2ab / 100 = 18

⇒ ab = 900

Now (a – b)² = (a + b)² – 4ab

= (100)² – [4 × 900]

= 10000 – 3600

= 6400

a – b = 80

Now, solving a + b = 100 and a – b = 80.

We get a = 90 and b = 10

Hence, the numbers are 90 and 10.

Question 2: Three numbers a, b, c are in GP. and ax = by = c^z, then prove that x, z will be in H.P.

Solution:

Let a^x = b^y = c^z = k

Then a = k^1/x , b = k^1/y , c = k^1/2 and a, b, c are in GP.

∴ b² = ac …(i)

Put the values of a, b, c in equation (i)

k^1/y = k^1/x * k^1/z

k^2/y = k^1/x+1/z which is only possible, when

2 / y = 1 / x + 1 / z

This shows that [1 / x], [1 / y], [1 / z] are in A.P.

Then x, y, z will be in H.P.

Question 3: Three numbers a, b, c are in H.P. Prove that 2a – b, b, 2c – b will be in GP.

Solution:

If a, b, c are in HP, then

2a – b, b, 2c – b will be in GP.

If b² = (2a – b) (2c – b)

b² = 4ac – 2bc – 2ab + b²

4ac = 2bc + 2ab

Divide by 2abc on both sides,

4ac / 2abc = 2bc / 2abc + 2ab / 2abc

It is clear that from equation (i) and (ii), if a, b, c are in H.P. then 2a – b, b, 2c – b will be in G.P.

Question 4: If a, b, c are in A.P., b, c, d are in GP. and c, d, e are in H.P., then prove that a, c, e will be in H.P.

Solution:

∵ a, b, c are in A.P.

b = [a + c] / [2] —- (1) and

b, c, d are in GP then

c² = bd —- (2)

Again c, d, e are in HP,

d = [2ce] / [c + e] —- (3)

Put the value of b and d from (1) and (3) in equation (2),

c² = bd

c² = [a + c] / [2] * [2ce] / [c + e]

c (c + e) = e (a + c)

c² = ae

So, a,c and e are in GP.

Question 5: If three numbers a, b, c are in A.P. and H.P. both then prove that they will be in GP. also.

Solution:

a, b, c will be in A.P. and H.P.

b = [a + c] / [2] —- (1)

b = [2ac] / [a + c]

= [ac] / {[a + c] / [2]}

= ac / b

b * b = ac

b = √ac

Hence a, b and c are in GP.