## Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Miscellaneous Exercise

Question 1.

10^{th} term of series – 4, – 1, + 2, + 5, … is :

(a) 23

(b) – 23

(c) 32

(d) – 32

Solution:

-4, -1, + 2, + 5, …

Here a = – 4

d = (- 1) – (- 4) = – 1 + 4 = 3

10th term = T1o

= a + 9d

= -4 + 9 × 3

= – 4 + 27 = 23

Hence, option (a) is correct.

Question 2.

9^{th} term of A.P. is 35 and 19th term is 75, then 20th term will be :

(a) 78

(b) 79

(c) 80

(d) 81

Solution:

Given,

T_{9} = 35

⇒ a + 8d = 35 …(i)

and T_{19} = 75

⇒ a + 18d = 75 …(ii)

Subtracting equation (i) from equation (ii),

10d = 40 ⇒ d = 4

Put d = 4 in equation (i),

a + 8 × 4 = 35

⇒ a + 32 = 35

⇒ a = 35 – 32 = 3 ⇒ a = 3

Thus, 20^{th} term = T_{20}

= a + 19 d

= 3 + 19 × 4 = 3 + 76 = 79

Hence, option (b) is correct.

Question 3.

Sum of n terms of series 1, 3, 5, … is :

(a) (n – 1)^{2}

(b)(n + 1)^{2}

(c) (2n – 1)^{2}

(d) n^{2}

Solution:

1,3,5, …

nth term of the given series

T_{n} = 2n – 1

Sum of n terms

s_{n} = ∑(2n – 1)

= 2∑n – ∑1

= 2 \(\frac { n(n+1) }{ 2 } \) – n

= n^{2} + n – n = n^{2}

Hence, option (d) is correct.

Question 4.

If first term of A.P. is 5, last term is 45 and sum of terms is 400, then numbers of terms is :

(a) 8

(b) 10

(c) 16

(d) 20

Solution:

Given, a = 5, l = 45, s_{n} = 400

Hence, option (c) is correct.

Question 5.

If 3^{rd} term of A.P. is 18 and 7th term is 30 then sum of first 17 terms will be :

(a) 600

(b) 612

(c) 624

(d) 636

Solution:

Given T_{3} = a + 2d

⇒ a + 2 d = 18 …(i)

and T_{7} = a + 6d

a + 6d = 30 …(ii)

From equation (i) and (ii),

4d = 12 ⇒ d = 3

Putting d = 3 in equation (i),

a + 2 × 3 = 18

⇒ a + 6 = 18

⇒ a = 12

Hence, option (b) is correct.

Question 6.

If (x + 1), 3x, (4x + 2) are in A.P., then 5^{th} term will be :

(a) 14

(b) 19

(c) 24

(d) 28

Solution:

Common difference,

d = 3x – (x + 1)

= 3 × 3 – (3 + 1)

= 9 – 4 = 5

∴ 5^{th} term = a + 4d

= 4 + 4 × 5

= 4 + 20 = 24

Hence, option (c) is correct.

Question 7.

a, b, c are in A.P., A.M. of a and b is x, A.M. of b and c is y, then A.M. of x and y will be :

(a) a

(b) b

(c) c

(d) a + c

Solution:

a, x, b, y, c are in AP., A.M. of x and y is b

Hence option (b) is correct.

Question 8.

Sum of n terms of A.P. is 3n^{2} + 5n its 27th term is:

(a) 160

(b) 162

(c) 164

(d) 166

Solution:

If Sn – 3n^{2} + 5n

Then, n^{th} term

T_{n} = s_{n}– s_{n-1}

= (3n^{2} + 5n) – [3(n – 1)^{2} + 5(n – 1)]

= (3n^{2} + 5n) – [3(n^{2} + 1 – 2n) + 5n – 5]

= 3n^{2} + 5n – [3n^{2} + 3 – 6n + 5n – 5]

= 3n^{2} + 5n – 3n^{2} + n + 2

= 6n + 2

Put n = 27

T_{27} = 6 × 27 + 2

= 162 + 2 = 164

Hence, option (c) is correct.

Question 9.

Sum of 50 A.M. between 20 and 30 is :

(a) 1255

(b) 1205

(c) 1250

(d) 1225

Solution:

First term, a = 20

Last term, l = 30

Number of terms = 52

Sum of 52 terms

S_{52} = \(\frac { 52 }{ 2 } \) (20 + 30)

= 26 × 50 = 1300

Thus, sum of 50 A.M. between 20 and 30 is

= 1300 – 20 – 30

= 1300 – 50 = 1250

Hence, option (c) is correct.

Question 10.

Common ratio of GP.

Solution:

Hence, option (a) is correct.

Question 11.

Number of terms in GP. 96, 48, 24,12, … \(\frac { 3 }{ 16 } \) is –

(a) 8

(b) 10

(c) 12

(d) 15

Solution:

Question 12.

Value of 9^{1/3} × 9^{1/9} × 9^{1/27} × … ∞-

(a) 1

(b) 3

(c) 9

(d) 27

Solution:

Question 13.

Sum of infinite terms of series

Solution:

Question 14.

Sum of infinite terms of series

Solution:

Question 15.

If third term of GP. is 2, then product of its first five terms is :

(a) 4

(b) 16

(c) 32

(d) 64

Solution:

Given : Let first five terms of GP. are

Question 16.

For which value of n, expression

will be GM. between a and b :

(a) 1

(b) 2

(c) 0

(d) – \(\frac { 1 }{ 2 } \)

Solution:

Or a^{2n+2} + b^{2n+2} + 2a^{n+1} b^{n+1}

= ab [a^{2n} + b^{2n} + 2a^{n}b^{n}]

Or a^{2n} a^{2} + b^{2n} b^{2} + 2ab.a^{n}b^{n} = ab[a^{2n} + b^{2n} + 2a^{n}b^{n}]

Or a^{2n}a^{2} – ab.a^{2n} + b^{2n}b^{2} – ab.b^{2n}

+ 2ab.a^{n}b^{n} – 2ab.a^{n}b^{n} = 0

Or a^{2n} a(a-b) + b^{2n }b(b-a) = 0

⇒ a^{2n+1} (a-b) – b^{2n+1} (a-b) = 0

Or (a-b) [a^{2n+1} – b^{2n+1}] = 0

Or a^{2n+1} = b^{2n+1}

[if a ≠ b ⇒ a – b ≠ 0]

Or \(\frac { { a }^{ 2n+1 } }{ { b }^{ 2n+1 } } \) = 1

Comparing powers of both sides

2n + 1 = 0

Or n = – \(\frac { 1 }{ 2 } \)

Thus, value of n = – \(\frac { 1 }{ 2 } \)

Hence option (d) is correct

Question 17.

If G_{1} and G_{2} are two GM between a and b then value of G_{1}G_{2} is :

(a) \(\sqrt { ab }\)

(b) ab

(c) (ab)^{2}

(d) (ab)^{3}

Solution:

Question 18.

GM. between – 9 and – 4 :

(a) – 36

(b) 6

(c) – 6

(d) 36

Solution:

Let GM. between – 9 and – 4 is G then

Question 19.

Series \(\frac { 1 }{ 2 } \),\(\frac { 5 }{ 13 } \),\(\frac { 5 }{ 16 } \), …. is

(a) A.P.

(b) GP.

(c) H.P.

(d) Other

Solution:

Question 20.

6th term of series 1,\(\frac { 1 }{ 4 } \),\(\frac { 1 }{ 7 } \),\(\frac { 1 }{ 10 } \),….. is:

(a) \(\frac { 1 }{ 13 } \)

(b) \(\frac { 1 }{ 16 } \)

(c) \(\frac { 1 }{ 15 } \)

(d) None of these

Solution:

The given series is a H.P. because the corresponding A.P. will be 1, 4, 7, 10,…. In which

a = 1, d = 4 – 1 = 3

∴ If 6th term a_{6} = a + 5d

= 1 + 5 × 3 = 1 + 15 = 16

Thus, 6th term of corresponding H.P. is \(\frac { 1 }{ 16 } \).

Hence, option (b) is correct.

Question 21.

If a, b, c,d are in H.P., then true statement is

(a) ab > cd

(b) ac > bd

(c) ad > be

(d) None of these

Solution:

Question 22.

If H.M. of two numbers is 4, A.M. and GM. is G if 2A + G^{2} = 27, then numbers are :

(a) 6, 4

(b) 8,2

(c) 8, 6

(d) 6, 3

Solution:

solving equations (i) and (ii), we get a = 6, b = 3

Hence, option (d) is correct.

Question 23.

If ratio of H.M. and GM. of two numbers is 12 : 13, then ratio of numbers will be :

(a) 1 : 2

(b) 2 : 3

(c) 3 : 5

(d) 4 : 9

Solution:

Let two numbers are a and b. Then its H.M.

Question 24.

If A, G, H are A.M., GM. and H.M. respectively, between two numbers a and b then A, G, H, will be :

(a) In H.P.

(b) In GP.

(c) In A.P.

(d) None of these

Solution:

Relation in A, G, H is G = \(\sqrt { AH }\)

G^{2} = AH

From this, it is clear that, A, G H will be in GP.

Hence, option (b) is correct.

Question 25.

If H be H.M. between numbers a and b, then value of \(\frac { H }{ a } \) + \(\frac { H }{ b } \) is :

(a) 2

(b) \(\frac { a+b }{ ab } \)

(c) \(\frac { ab }{ a+b } \)

(d) None of these

Solution:

Question 26.

If a, b, c are in H.P., then correct statement is –

Solution:

H.M. of a, b, c

H = \(\frac { 2ac }{ a+c } \) ….(i)

G.M. of a,b,c

G = \(\sqrt { ac }\) ….(ii)

From equation (i) and (ii),

b = \(\frac { 2ac }{ a+c } \)

We know that GM. > H.M.

G > H

\(\sqrt { ac }\) > b

Hence, option (c) is correct

Question 27.

If nth term of any series is \(\frac { { n }^{ 2 } }{ { 3 }^{ n } } \) then write first 3 terms of series.

Solution:

Question 28.

Which term of progression 72,70,68, 66, is …….. 40 ?

Solution:

Here a = 72, d = 70 – 72 = – 2

Let nth term is 40, then

T_{n} =40

⇒ a + (n – 1) d = 40

⇒ 72 + (n – 1) × (- 2) = 40

⇒ – 2 (n – 1) = 40 – 72

⇒ -2 (n-1) = -32

⇒ n – 1 = 16

n = 17

Hence, 17th term of the progression is 40.

Question 29.

If in an A.P., sum of m and n terms are in ratio m^{2} : n^{2}, then prove that ratio of mth and nth term will be (2m – 1) ; (2n – 1).

Solution:

Let first term of A.P. is a and common difference is d, then

Question 30.

If sides of any right angled triangle are in A.P., then find the ratio of length of their sides.

Solution:

Let sides of right angled triangle are

a – d, a, a + d

By Pythagoras theorem

(a + d)^{2} – (a – d)^{2} + a^{2}

(∵ hypotenuse is longest side)

Question 31.

– \(\frac { 2 }{ 7 } \), a, – \(\frac { 7 }{ 2 } \) are in G.P., then find value of a.

Solution:

Question 32.

Find sum of n terms of series 1 – 1 + 1 – 1 + …..

Solution:

Question 33.

Find the value of 3^{2}.

2^{1/2}**.**4^{1/8}**.**16^{1/32}…∞

Solution:

Question 34.

For which value of n, expression ?

will be H.M. of two numbers a and b ?

Solution:

Question 35.

If A and Hare A.M. and H.M. between a and b, then prove that

Solution:

Question 36.

If a, b, c are in A.P. and b, c, d are in H.P., then prove that ad = bc.

Solution:

Question 37.

If a + b … + l are in GP., then prove that its

Solution:

Question 38.

Find the sum of n terms of sequence 3, 33, 333,… .

Solution:

It is clear that the given series is not in GP. but it can be related to GP. by writting it in the following form

Question 39.

Find the sum of sequence made by product of corresponding terms of sequence 1, 2, 4, 8, 16, 32 and sequence

32, 8, 2, \(\frac { 1 }{ 2 } \), \(\frac { 1 }{ 8 } \), \(\frac { 1 }{ 32 } \)

Solution:

Sequence made by product of corresponding terms of sequence 1, 2, 4, 8, 16, 32 and sequence 32, 8, 2,

Question 40.

Find the value of n so that

is G.M. between a and b.

Solution:

Question 41.

If G_{1} and G_{2} are two geometric mean between a and b, then prove that G_{1},G_{2} = ab

Solution:

Question 42.

If arithmetic mean (A.M.) and geometric mean (GM.) of any two numbers a and b are in ratio m : n, then prove that

Solution:

Question 43.

A.M. of two numbers is 50 and H.M. is 18,find the numbers.

Solution:

Let a and b are two numbers, then

Question 44.

The difference between A.M. and GM. of two numbers is 2, difference between GM. and H.M. is 1.2. Find the numbers.

Solution:

According to question,

Question 45.

If a, b, c are in A.P., x,y, z are in H.P. and ax, by, cz are in GP. then prove that:

Solution:

Question 46.

A_{1},A_{2}, are two A.M. between two positive numbers a and b two GM. G_{1}, G_{2} and two H.M. H_{1}, H_{2}, then prove that :

A_{1}H_{2} = A_{2}H_{1} = G_{1}G_{2} = ab

Solution:

Let two numbers are a and b, then

A_{1}, A_{2} are two A.M. between a and b, then

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