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

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

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.1

Question 1.
From the following sequence 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 common difference of two consecutive terms is same.

Question 2.
Find first term, common difference and 5th term of those sequence which have following nth term :
(i) 3n+ 7
(ii) a + (n – 1)d
(iii) 5 – 3n.
Solution :
(i) 3n + 7
nth 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
nth 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
T₂ = 5 – 3 × 2 = 5 – 6 = -1
Common difference
d = T₂ – T₁= a + d – a = d
Fifth term T₅ = a + (5 – 1)d = a + 4d
Hence, ‘a’ = a,’d’ = d and T₅ = a + 4d

(iii) 5 – 3n
nth 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 sequence of following nth terms, are not in A.P. :

Solution :
If the value of T_n +1 – T_n is not independent of n then sequence of nth terms are not in A.P.

Hence, given sequence of nth term is not in
(ii) T_n+1-T_n = {(n+ 1)² + 1} – {n² + 1}
= (n² + 1 + 2n + 1) – (n² + 1)
= (n² + 2n + 2 – n² – 1)
= 2n+ 1

Question 4.
In an A.P. 2 + 5 + 8 + 11 +… which term is 65 ?
Solution :
Given series = 2 + 5 + 8 + 11 + ….
Let its nth term is 65
Then T_n = 65 and d = 5 – 2 = 3
⇒ a + (n – 1)d – 65
⇒ 2 + (n – 1) × 3 = 65
⇒ 2 + 3n – 3 = 65
⇒ 3n – 1 = 65

Hence, 22nd term is 65

Question 5.
In an A.P. 4 + 9 + 14 + 19 +… + 124, find 13th term from last.
Solution :
Given series = 4 + 9 + 14 + 19 +…+ 124 a = 4, d = 9 – 4 = 5, l= 124
13th term from last
= l – (n – 1)d
= 124 – (13 – 1) × 5
= 124 – 12 × 5
= 124 – 60 = 64
Hence, 13th term from last is 64.

Question 6.
In an A.P. 2 + 5 + 8 + 11 + … if last term is 95, then find the number of terms of the series.
Solution:
Given series = 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) = 39/3
= 31
⇒ n = 31 + 1 = 32
Hence, number of terms of the series is 32.

Question 7.
If 9th term of A.P. is zero, then prove that 29th term is twice the 19th term.
Solution :
Let a be the first term and d be the common difference then 9th
term = T₉ = 0 Then a + (9 – 1)d= 0
⇒ a + 8d = 0   ….. (i)
29th term = T₂₉ = a + 28d  …(ii)
19th term = T₁₉ = a + 18d  …(iii)
Putting the value of equation (i) into equation (ii) and (iii)
we get
T₂₉ = (a + 8 d) + 20 d
⇒ T₂₉ = 0 + 20 d
⇒ T₂₉ = 20d … (iv)
⇒ T₂₉ = (a + 8 d) + 10 d
⇒ T₂₉ = 0 + 10d
⇒ T₂₉ = 10d … (v)
From equation (iv) and (v), we have
T₂₉ = 20d = 2 × 10d =2 × T₂₉
T₂₉ = 2T₂₉
Hence Proved.

Question 8.
How many two digit natural number which are divisible by 3 ?
Solution :
First two digit number which is divisible by 3 is 12 and last two digit number is 99
∵ We have to find only the numbers which are divisible by 3, so 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, two digit natural number which are divisible by 3 are 30.

Question 9.
If in A.P., pth term is q and qth term is p, then find (p + q)th term.
Solution :
Let first term of A.P. = a
and common difference = d
Then pth term = T_p = a + (p – 1)d = q
⇒ a + (p – 1)d = q … (i)
and qth term =  T_p = a + (q – 1)d = p
⇒ a + (q – 1)d = p   …. (ii)
(p + qth 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. pth term is 1/q and 17th term 1/p, then prove that pqth term is unity.
Solution :
Let first term be a and common difference be d

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.2

Question 1.
Find the sum of the following progression :

Solution:
(i) Given progression
7 + 11 + 15 + 19 + …
Here, a = 7, d = 11 – 7 = 4, n = 20
Then S_n =  n/2
S₂₀ = 20/2
= 10 [14 + 19 × 4]
= 10 × (14 + 76)
= 10 × 90 = 900
Hence, sum of 20 terms is 900.

Question 2.
Find the sum of 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 number at last term is n, then last term
= 3 + (n – 1) × 6 = 99
[From formula, l = a + (n + 1 )d]
⇒ 6 (n – 1) = 99 – 3
⇒ (n – 1) = 96/3
=16 6
⇒ n = 16 + 1 = 17
Hence, sum of 17 terms,

= 17 × 51 = 867
Hence, 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 rth term is 2r + 3.
Solution:
Given,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,

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

Question 4.
Sum of n terms of any A.P. is n² + 2n. Find the first term and common difference.
Solution:
Sum of terms
S_n= n² + 2n
and S, (n – 1)² + 2(n-1)
We know that nth term of A.P.
T_n= S_n – S_{n – 1}
⇒ T_n= (n² + 2n) – [(n – 1)² + 2(n − 1)]
= [n² + 2n] – [n² + 1 – 2n + 2n – 2]
= [n² + 2n] – [n² – 1]
= n² + 2n – n² + 1
= 2n + 1
First term T₁= 2 × 1 + 1 = 2 + 1 = 3
Second term, T₂ = 2 2 + 1 = 4 + 1 = 5
Common difference d = T₂ – T₁ = 5 – 3 = 2
Hence, first term is 3 and common difference is 2.

Question 5.
If sum of n terms of A.P. 1, 6, 11,…. is 148, then find 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,

Taking ‘+’, n = 8
Taking ‘-‘, n = –37/5
Negative value of n is impossible.
Hence, n = 8
Last term= T₉ = a + 7d
= 1 +7 × 5 = 1 + 35 = 36
Hence, number of terms is 8 and last term is 36.

Question 6.
If in an A.P„ sum of p terms is equal to sum of q terms, then find the sum of (p + q) terms.
Solution:
Let a₁ is the first term and d is the common difference of A.P.
Then, sum of first p terms

and sum of first q terms

According to question,
some of first p terms = sum of first q terms


Now, sum of (p + q) terms of this progression

Hence, sum of (p + q) terms of given A.P. is zero.

Question 7.
If 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 is first term and dis common difference of an A.P. then

Subtracting equation (i) from equation (ii)

Multiplying by 3 in both sides

= S₃  [From equation (iii)]
Hence, 3 (S₂ – S₁) = S3
Hence Proved.

Question 8.
If sum of m A.P. of n terms are S₁, S₂ and S₃…, S_m respectively. Their first term are 1, 2, 3,…., m respectively and common difference 1, 3, 5,…., (2m – 1) respectively, then prove that
S₁ + S₂ + S₃ + … + S_m= mn/2 (mn + 1)
Solution:

Question 9.
If sum of first p, q, r terms of any A.P. are a, b, c respectively, then prove that :

Solution:
Let a is the first term and d is the common difference of the given A.P. Then according to question,

From equation (i),(ii) and (iii)

Now, multiplying equation (iv), (v) and (vi) by (q – r), (r – p) and (p – q) respectively.

Adding equations (vii), (viii) and (ix),

Question 10.
Find three numbers in A.P., whose sum is 12 and sum of their cube is 408.
Solution:
Let three numbers are a – d, a and a + d
Then (a – d) + (a) + (a + d) = 12
⇒ 3a = 12 or a = 4
According to question,
(a – d)³ + a³ + (a + d)³ = 408
⇒ (4 – d)³ + (4³) + (4 + d)³ = 408
⇒ (4³ – 3(4²)d + 3 × 4 × d² – d³) + (4³)
+ (4³ + 3(4²)d + 3 × 4 × d² + d³) = 408
⇒ 3(4³) + 24d² = 408
⇒ 24d² = 408 – 192 = 216
⇒ d² =  216/24
⇒ d² = 9
⇒ d = 3

Question 11.
If n arithmetic mean are inserted in between 1 and 51 such that ratio of 4th and 7th arithmetic mean is 3 : 5 dthen find the value of n.
Solution:
Let n arithmetic mean between 1 and 51 then
1, A₁ + A₂ + A₃ + …, A_n 51
(n + 2)th term = 51=1
Then 51 = 1 + (n + 2 – 1)d
[From formula l = a + (n – 1)d]
⇒ (1 + n)d = 51 – 1 = 50
⇒ (n + 1)d = 50

⇒ 5n + 5 × 201 = 3n + 3 × 351
⇒ 5n – 3n = 1053 – 1005
⇒ 2n = 48
⇒ n = 24
Hence, n = 24

Question 12.
If x, y, z are in A.P., then prove that :

Solution:
(i) y + z, z + x, x + y will be in A.P. if
⇒ (z + x) – (y + z) = (x + y) – (z + x)
⇒ z + x – y – z = x + y – z – x
⇒ z + x = y – z
⇒ z + x = 2y
⇒ y = x+z/2
Hence, x, y, z are in A.P., which are given.
Hence, y + z, z + x, x + y are in A.P.
Hence Proved.

Hence, x,y,z are in A.P., which are given.

Hence Proved.

Question 13.
If x² (y + z), y² (z + x), z²(x + y) are in A.P., then prove that either x, y, z are in A.P. or xy + yz + zx = 0.
Solution:
∵ x²(y + z), y²(z + x), z²(x + y) are in A.P.
∴ Adding xyz in each terms x²(y + z) + xyz, y²(z + x) + xyz, z²(x + y) + xyz also will be in A.P.
or x(xy + yz + zx), y(xy +yz + zx), z(xy + yz + zx) also will be in A.P.
∴ 2y(xy + yz + zx) = x(xy + yz + zx) + z(xy + yz + zx)
⇒ 2y(xy + yz + zx) = (xy + yz + zx) (x + z)
⇒ 2y(xy + yz + zx) – (xy + yz + zx) (x + z) = 0
⇒ (xy + yz + zx) (2y – x – z) = 0
If  2y – x – z = 0
Then 2y = x + z
⇒ x, y, z are in A.P.
or xy + yz + zx = 0
Hence Proved.

Question 14.
Find the sum of A.P. a₁ + a₂ + a₃ …, A₃₀.
Given that
a₁ + a₇ + a_10 + a₂₁ + a₂₄ + a_{30 = 540}
Solution:
Number of terms in series = 30, we know that sum of same distant terms from start and end remains constant and is equal to sum of first and last term, i.e.,
^Tr ^{+T_{r – 1}} = a + 1
∵ a₇ is 7th term from start and a₂₄ is 7th term from last
a₇ + a₂₄ = a₁ + a₃₀  …(i)
Similarly, a₁₀ is 10th term from start and a₂₁ is 10th term from last, 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 = 30/2 (a₁ + a₃₀)
= 15 (180) = 2700
Hence, a₁ + a₂ + a₃ + …….+ a₃₀ = 2700

Question 15.
Interior angles of a polygon are in A.P. Smallest interior angle is 52° and difference at consecutive interior angles is 8°, then find number of sides of the polygon.
Solution:
Smallest angle = 52°
Difference of consecutive angles = 8°
Let number of sides of polygon be x.
First term a – 50
Common difference d = 8°
We know that, 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 dⁿ = a + (n + 1 )d
= 52° + (30 – 1)8°
= 52° + 29 × 8°
= 52° + 232°
= 284°
which is impossible
Hence, number of sides is 3.

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.3

Question 1.
(i) Find the 7^th term of series 1 + 3 + 9 + 27 + …
(ii) Find the 10th term of

Solution:
(i) Given series = 1 + 3 + 9 + 27 + ….
Hedre, 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.
(i) Which term of series 64 + 32 + 16 + 8 + … is 1/64 ?
(ii) Which term of series 6 + 3 + 3/2 + 3/4 + … is 3/256 ?
Solution:
(i) Given series =64 + 32 + 16 + 8 +….

Question 3.
Find the common ratio and 12^th term of G.P. 5 + 10 + 20 + 40 + …
Solution:
Given
series = 5 + 10 + 20 + 40 + ….

n^th term = T_n = ar^{n – 1}
= 5 × (2)^{n – 1} = 5.2^{n – 1}

Question 4.
Find the fifth term from the last term of G.P. 2, 6, 18, 54, … 118098.
Solution:
Given
series = 2, 6, 18, 54,…..,118098
Common ratio r = 6/2 = 3
Last term = 118098
⇒ ar^{n – 1} = 118098
⇒ 2 × (3)^{n – 1} = 118098
⇒ (3)^{n – 1} = 59049 = (3)¹⁰
On comparing, n – 1 = 10
⇒  n = 11
5^th term from last =ar^{n – 5}
= 2 × (3)^{11 – 5} =2 × (3)⁶
= 2 × 72 = 1458

Question 5.
Third term of a G.P. is 32 and 7th term is 8192, then find 10th term of GP.
Solution:
Given, T₃= ar² = 32
T₇ = ar⁶ = 8192
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
10th term = T₁₀ = ar⁹
= 2 × (4)⁹ = 524288

Question 6.
Find G.P., whose 3^rd term is 1 and 7^th term is 16.
Solution:
Let a is first term of G.P. and r is common ratio, then
Third term T₃ = 1
⇒ ar²   = 1
Seventh term T₇ = 16
ar⁶ = 16
Divide equation (ii) from (i), we get

⇒ r⁴ = 16 = 2⁴
Hence, r = 2
ar² = 1
a × (2)² = 1

Question 7.
(i) Find 3 G.M. between 3 and 48.
(ii) Find 6 G.M. between 2 and 256.
Solution:
(i) Let G₁, G₂, G₃ be three GM. between 3 and 48, then
3, G₁, G₂, G₃, 48 will be in G.P.
Here a = 3,
Last term (5th 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, three G.M. are 6, 12, 24.

(ii) Let G₁, G₂, G₃, G₄, G₅, G₆ be six G.M. between 2 and 256, then
2, G₁, G₂, G₃, G₄, G₅, G₆, 256 will be in G.P.
Here a = 6, last term (8th term) = 256 and n = 8
Hence, 2.r⁷ = 256 [∵ l = ar^{n – 1}]
⇒ r⁷= 128 = 2⁷
⇒ r = 2
G₁ = 2 × 2 = 4
G₂ = 4 × 2 = 8
G₃ = 8 × 2 = 16
G₄ = 16 × 2 = 32
G₅ = 32 × 2 = 64
G₆ = 64 × 2 = 128
Hence six G.M. are 4, 8, 16, 32, 64, 128.

Question 8.
For which value of x, numbers x, x + 3, x + 9 are in G.P. ?
Solution:
x, x + 3, x + 9 are in G.P.
Then, (x + 3)² = x × (x + 9)
⇒ x² + 9 + 6x = x² + 9x
⇒ x² + 6x + 9 – x² – 9x = 0
⇒ – 3x + 9 = 0
⇒ −9/−3 = 3
Hence, x = 3

Question 9.
Find four terms which are in G.P., whose third term is 4 more than first term and second term is 36 more than 4^th term.
Solution:
Let first four terms of G.P. are a, ar, ar². ar³ where a is first term and r is common ratio.
According to question,
Third term = First term + 4
⇒ ar² = a + 4
⇒ a(r² – 1) = 4
According to question,
Second term = Fourth term + 36
⇒ ar = ar³ + 36
⇒ ar – ar³ = 36
⇒ ar(1 – r²) = 36
Divide equation (ii) by (i), we get

⇒ – r = 9
⇒ r = – 9
Hence, common ratio of G.P.,
r = – 9
Putting r = – 9 in equation (i),
a[(-9)² – 1] = 4
⇒ a (81 – 1) = 4

Question 10.
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 first term and R is common ratio of G.P., then according to 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 = a² R¹² Hence, from equation (iv) and (v)
q² = pr         ….. (v)
Hence Proved.

Question 11.
If (p + q)^th term in GP. is x and (p – q)^th term is y, then find p^th term.
Solution:
Let a is first term and r is common ratio of G.P.
So, (p + q)^th term = ar^p + q – 1 = x   … (i)
and (p – q)^th term = ar^p – q – 1 = y   … (ii)
Multiplying equation (i) and (ii), we get
a² (r^p + q – 1 + p – q) = x × y
or a² r^2p-2 = xy
ora² r^{2(p – 1)} = xy
or (ar^{p – 1})² = xy
or ar^{p – 1} = √xy
Hence, pth term = ar^{p – 1}
= √xy

Question 12.
If a, b, c are in G.P. and  a^x = b^y = c^z , then prove that

Solution:
Let a^x= b^y = C^z = k

and a, b, c are in G.P.
b² = ac
Putting the value of a, b, c in equation (i),

Question 13.
If n G.M. inserted between a and b then prove that 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

Question 14.
If x, y, z are in G.P., arithmetic mean of x, y is A₁ and A.M. of y, z is A₂ then prove that :

Solution:
x, y, z are in G.P.
∴ y² = xz     …… (i)
Arithmetic mean of x and y is A₁

Arithmetic mean of y and z is A₂

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.4

Question 1.
Find the sum of the following geometric progression :

Solution:
(i) Given progression
= 2 + 6 +18 + 51 +….

Question 2.
Find the sum of following geometric progression :
(i) 2 + 6 + 18 + 54 + … + 486
(ii) 64 + 32 + 16 + … + 1/4
Solution:
(i) Given progression
= 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

s₆ = 3⁶ – 1
= 729 -1 = 728
Hence, sum is 728.
(ii) Given progression

Question 3.
Sum of how many terms of G.P. 4, 12, 36, is 484 ?
Solution:
a = 4, r = 12/4 = 3, S_n = 484, (r >1)
S_n = 484

3ⁿ – 1 = 242
3ⁿ = 242 + 1 = 243
⇒  3ⁿ = 3⁵
Hence, number of term is 5.

Question  4.
Sum of first 5 terms of any G.P. is 124 and common ratio is 2. Find the first term of the series.
Solution:
n = 5, S_n = 124, r = 2, r > 1
S_n = 124

Hence, first term is 4.

Question 5.
The common ratio of any G.P. is 2, last term 160 and sum is 310. Find the first term of the series.
Solution:
r = 2,T_n= 160, S_n = 310, r > 1,
ar^{n – 1} = 160
⇒ a(2)^n – 1 = 160

⇒ a(2)ⁿ – a = 310    … (ii)
From equation (i) and … (ii)
a = 320 – 310 = 10
Hence, first term is 10.

Question 6.
Find the sum of first n terms of the following series :
(i) 7 + 11 + 777 + …
(ii) .5 + .55 + .555 + ….
(iii) .9 + .99 + .999 + ….
Solution:
(i) Given series upto n terms 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]

(ii) 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 =.5 + .55 + .555 + .5555 + ….upto n terms]

Question 7.
Find the rational form of the following recurring decimals :
(i) 2.35¯
(ii) .625¯
(iii) 2.752¯
Solution:
(i) Digit 5 is repeated in 2.35 .
∴ 2.35 = 2.35555…………
= 2 + [.3 + .05 + .005 + .0005+ … upto infinity]

Question 8.
First term of any infinite series is 64 and each term is three times its succeeding terms. Find the series.
Solution:
Let infinte series is
a + ar + ar² + ar^-1 + …
Given, First term = 64
According to question,
a = 3 (ar + ar² + ar³ + … ∞)
⇒ a = 3ar(1 + r + r² + … ∞)
⇒ 1 = 3r (1/1−r).
⇒ 1 – r = 3r
⇒ 4r = 1 or r = 1/4

Question 9.
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)

Question 10.
If x = 1 + a + a² + … ∞, where | a | < 1 and y = 1 + b + b² + … = ∞, where | b | < 1, then prove that

Solution:
Given x = 1 + a + a² + … ∞ (| a | < 1)

Question 11.
Find the sum of the series

Solution:
Given series

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.5

Question 1.
Find the sum of n terms of following series:

Solution:

Let sum of this series is S_n

Hence,arithmetic geometric series

(ii) 1 + 3x + 5x² + 7x³ + …
In the given series 1, 3, 5, 7, … are in A.P. and 1, x, x², … are in GP.
The nth term of A.P. is (2n – 1) and nth term of G.P. is x^{n – 1} .
Hence, nth term of arithmetic-geometric series will be (2n – 1)x^{n – 1} .
Let sum of n terms of arithmetic-geometric series is S_n then
S_n = 1 + 3x + 5x² + 7x³ + … (2n – 1)x^{n – 1} … (i)
Multiplying on both sides by x
xS_n = x + 5x² + 3x³ + … [2(n – 1) – 1 ]x^{n – 1} 5(2n – 1)xⁿ … (ii)
Subtracting equation (ii) from (i), we get
(1 – x)S_n = (1 + 2x + 2x² + 2x³ + … + 2x^{n – 1}) + (2n – 1)xⁿ

Hence, it is sum of n terms.

Let sum of its n terms is S_n then

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

Question 2.
Find the sum of infinite terms of following series :

Solution:
(i) Let sum of the given series is S_∞ then

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

Hence, sum of the series is 6/7.
(ii) Let sum of the given series is S_∞ then

Adding equation (i) and (ii), we get

Hence, sum of the series is 3/16.
(iii) Let sum of the given series is S_∞ then
S_∞ = 2x + 3x² – 4x³
S_∞ = x – 2x² + 3x³
Adding equation (i) and (ii), we get
(1 + x)S_∞ x – 2x ² 3x ³ +…

Question 3.
Find n^th term and sum of n terms of following series :
(i )2 + 5 + 14 + 41 + 122 + …
(ii) 3.2 + 5.2² + 7.2³ + …
(iii) 1 + 4x + 7x² + 10x³ + …
Solution:
(i) Difference of consecutive terms of given series 3,9, 27,… are in G.P.
Let sum of its n terms is S_n and n^th term is T_n , then
S_n  =2 + 5 + 14 + 41 + 122 + … + T_n … (i)
By increasing on place
S_n  =2 + 5+ 14 + 41 + 122 + … + T_{n – 1} + T_n … (ii)
Subtracting equation (ii) from (i), we get
0 = 2 + (3 + 9 + 27 + … + (n – 1) term) – T_n
⇒ T_n = 2 + (3 + 9 + 27 +… + (n- 1) term)

Let the sum of series is S_n on putting the value of n = 1, 2, 3, …, then

(ii) In the given series, A.P. is 3, 5, 7, … and its n ^th term
= 3 + (n – 1)2 = 3 + 2n – 2 = 2n + 1. and G.P.
is 2, 2², 2³, … and its n^th term is 2ⁿ.
Let sum of its n terms is S_n and n^th term is T_n ,
Then term of given arithmetic-geometric series is
T_n = (2n + 1)2ⁿ.
Let the sum of series is S_n on putting the value of n in this, then
S_n = 3.2 + 5.2² + 7.2³ + … +(2n – 1)2^{n + 1} + (2n+ 1)2ⁿ …(i)

2S_n = 3.2² + 5.2³ +…+ (2n – 1)2ⁿ + (2n + 1)2^{n + 1}…(ii)
Subtracting equation (ii) from (i), we get
-S_n = 3.2 + 2[2² + 2³ + …2ⁿ] -(2n+ 1)2^{n + 1}

⇒ -S_n = 6 + 2² (2^{n + 1} – 1) – (2n + 1)2^n + 1
⇒ -S_n = 6 + 2^{n + 2} – 8 – (2n + 1)2^{n + 1}
⇒ -S_n = 2ⁿ⁺¹ [2 – 2n – 1] – 2
⇒ -S_n = (2n- 1)2^{n + 1} + 2
Hence, sum of n terms
= (2n + 5)2^{n + 1} + 2.

(iii) In the given series, A.P. is 1, 4, 7, 10,… and its n^th term
= 1 + (n – 1)³ = 3n – 2 and G.P.
is 1,x, x², x³,… and its n^th term = x^{n – 1}
Hence, n^th term of arithmetic-geometric series is (3n – 2)x^{n – 1}and sum is S_n
than S_n= 1 + 4x + 7x² + 10x³ +…+ (3n – 5)x^{n – 2}
+ (3n – 2)x^{n – 1} … (i)
xS_n = x + 4x² + 7x³ +…+ (3n – 5)x^{n – 1}
+ (3n – 2)x^{n – 1} … (ii)
Subtracting equation (ii) from (i), we get
(1 – x)S_n = 1 + [3x + 3x² + 3x³ +. ..3x^{n – 1}] – (3n- 2)x^{n + 1}

Question 4.
Find the sum of n terms of series 2 + 5x + 8x² + 11x³ + … and hence obtained the sum of infinite series | x | < 1.
Solution:
In the given series, A. P. is 2, 5, 8, 11, … and its n^th term
= 2 + (n – 1 )3 = 2 + 3n – 3 = 3n – 1 and G.P
is 1, x, x², x³,… and its nth term is x^{n – 1} and sum of
arithmetic-geometric series be S_n and its term is (2n – 1)x^{n – 1} so

S_n = 2 + 5x + 8x² + 11x³ +…+ (3n – 4)x^{n – 2} + (3n – 1)x^{n – 1} … (i)

xS_n = 2x + 5x² + 8x³ + … + (3n – 4)x^{n –}  + (3n- 1)xⁿ … (ii)
Subtracting equation (ii) from (i), we get
(1 – x)S_n = 2 + [3x + 3x² + 3x³ + … + 3x^{n – 1}] + (3n – 1)xⁿ

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.6

Question 1.
Find the sum of n term of that series whose nth term is :
(i) 3n² + 2n + 5,
(ii) 4n³ + 7n + 1
(iii) n(n + 1)(n + 2)
Solution:


(iii) Here T_n = n (n + 1)(n + 2)
⇒ T_n = n(n² + 3n + 2)
⇒ T_n = n³ + 3n² + 2n
∴ S_n = ∑T_n
⇒ S_n = ∑n³ + 3∑n² + 2∑M

Question 2.
Find the sum of it terms of following series :
(i) 3² + 7² + 11² + 15² + …
(ii) 2³ + 5³ + 8³ + 11³ + …
(iii) 1.2² + 2.3² + 3.4² + …
Solution:
(i) 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)
⇒ Sn = 16∑n² – 8∑M + ∑1

(ii) 2³ + 5³ + 8³ + 11³ + …
n^th term of the given series
T_n = [2 + (n – 1)3]³
= (2 + 3M – 3)³ = (3n – 1)³
= (3n)³ – 3(3n)² × 1 + 3(3n) × 1² – 1³
= 27n³ – 27n² + 9n – 1
∴ S_n = ∑(27n³ – 27n² + 9n – 1)
⇒ s_n = 27∑n³ – 27∑n² + 9∑n – ∑1

(iii) 1.22 + 2.32 + 3.42 + …
n^th term of the given series
T_n = n.(n + 1)²
∴ S_n = ∑n(n + 1 )²
⇒ S_n = ∑[n(n² + 2n + 1)]
= ∑[n³ + 2 n² + n]
= ∑n³ + 2∑M² + ∑n

Question 3.
Find the n^th term and sum of n terms of following series :
(i) 1.3+3.5+ 5.7 + …
(ii) 1.2.4 + 2.3.7 + 3.4.10 + …
Solution:
(i) 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)
∴ S_n = ∑ (4n² – 1)
= 4∑n² – ∑1

(ii) 1.2.4 + 2.3.7 + 3.4.10 + …
n^th term of the given series
T_n = (1 + 2 + 3 + … n^th term)
(2 + 3 + 4 + … n^th term )
(4 + 7 + 10 + … n^th term)
= n . (n + 1) . [4 + (n – 1)³]
= n(n + 1) (3n + 1)
= (n² + n) (3n + 1)
= 3n³ + 3n² + n² + n
= 3n³ + 4n² + n = n(n + 1) (3n + 1)
∴ Sn = ∑(3n³ + 4n² + n)
= 3∑n³ + 4∑n² + ∑M

Question 4.
Find the n^th term and sum of n terms of the following series :
(i) 3 + 8 + 15 + 24 + …
(ii) 1 + 6 + 13 + 22 …
Solution:
(i) 3 + 8 + 15 + 24 + …
Let n^th term of the series is T_n and sum of n terms is s_n then
s_n = 3 + 8 + 15 + 24 + … + T_n
Again s_n = 3 + 8 + 15 + 24 + … + T_{n – 1} + T_n
On subtracting
0 = 3 + 5 + 7 + 9 + … n upto n terms – T_n
⇒ T_n = 3 + 5 + 7 + 9 + … n upto n terms

(ii) 1 + 6 + 13 + 22 +…
The difference of consecutive terms of given series 5, 7, 9, … are in A.P. So n^th term and sum of its n terms will be find by difference method.
Let n^th term of the series be Tn and sum of nth terms is S_n, then
S_n = 1 + 6 + 13 + 22 + … + T_n …(i)
By increasing one place
S_n = 1 + 6 + 13 + … + T_n-1 + T_n …(ii)
Subtracting (ii) from (i)

Question 5.
Find the nth term and sum of n term of the following series :
(i) 1 + (1 + 2) + (1 + 2 + 3) + …
(ii) 1² + (1² + 2²) + (1² + 2² + 3²) + …
Solution:
(i) 1 + (1 + 2) + (1 + 2 + 3) + …
Let n^th term of the series be Tn and sum of n terms is S_n, then

(ii) 1² + (1² + 2²) + (1² + 2² + 3²) + …
Let n^th term of the series be T_n and sum of n terms is s_n, then

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 = 1/179

Hence, 10^th term of corresponding H.P. is = 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 = 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 = 2/na+(2−n)b

Question 3.
Find that H.P. whose 2nd term is 2/5 and 7th term is 4/25
Solution:
Here 2^nd term of corresponding A.P. is 5/2 and 7th term is 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 2/17 and 2/13


Hence, 20th term of corresponding H.P. will be 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 1/16.
So, 1, H₁, H₂, H₃, H₄, 1/16 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 −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 pq/(p+q)
Solution:
Let p^th term of corresponding A.P. is 1/q and q^th term is 1/q 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 = 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:

Rajasthan Board RBSE Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.8

Question 1.
A.M. of two numbers is 60 and H.M. is 18, Find the numbers.
Solution:
Let two numbers are a and b, then
a+b/2 = 50 ⇒ a + b = 100
and 2ab/a+b = 18
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 = √6400 = 80
Now, solving a + b = 100 and a – b = 80.
We get a = 90 and b = 10
Hence, numbers are 90 and 10.

Question 2.
If ratio of H.M. and GM. of two numbers is
12 : 13, then prove that ratio of number is 4 : 9.
Solution:
Let two number are a and 6 then its

Question 3.
The difference between A.M. and GM. is 2, difference between GM. and H.M., is 1.2. Find the numbers.
Solution:
According to question
A – G = 2 and A = G + 2
and G – H = 1.2 and H = G – 1.2
We know that,
G = √AH
⇒ G² = (G + 2) (G – 1.2)
⇒ G² = G² + 2G – 1.2G – 2.4
⇒ 0.8G = 2.4
⇒ G = 3
∴ A = 5, G = 3, H = 1.8

∴ a – b = 8 …(iii)
Solving equation (i) and (iii)
a = 9 and b = 1
Hence, number are 9 and 1.

Question 4.
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 value of a, b, c in equation (i)
k^1/y = k^1/x, k^1/z or 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 5.
Three numbers a, b, c are in H.P. Prove that 2a – b, b, 2c – b will be in GP.
Solution:

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 6.
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 7.
A₁ A₂ are two A.M. between two positive numbers a and b, G₁, G₂ are two GM. and H₁, H₂ are two H.M. then prove that:
(i) A₁H₂ = A₂H₁ = G₁ G₂ = ab
(ii) G₁G₂ : H₁H₂ = (A₁ + A₂): (H₁ + H₂)
Solution:
Let two numbers are a and b.
Then A₁, A₂ are two A.M. between a and b.

Again, G₁, G₂ are two GM. between a and b.
Then, a, G₁, G₂, b will be in G.P.

Question 8.
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.

Put the value of b and d from (i) and (iii) in equation (ii),

Question 9.
If three numbers a, b, c are in A.P. and H.P.
both than prove that they will be in GP. also.
Solution:
a, b, c will be in A.P. and H.P.

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,
T₉ = 35
⇒ a + 8d = 35 …(i)
and T₁₉ = 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 = T₂₀
= 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) n²
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 n(n+1)/2 – n
= n² + n – n = n²
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 T₃ = a + 2d
⇒ a + 2 d = 18 …(i)
and T₇ = 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 3n² + 5n its 27th term is:
(a) 160
(b) 162
(c) 164
(d) 166
Solution:
If Sn – 3n² + 5n
Then, n^th term
T_n = s_n– s_n-1
= (3n² + 5n) – [3(n – 1)² + 5(n – 1)]
= (3n² + 5n) – [3(n² + 1 – 2n) + 5n – 5]
= 3n² + 5n – [3n² + 3 – 6n + 5n – 5]
= 3n² + 5n – 3n² + n + 2
= 6n + 2
Put n = 27
T₂₇ = 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
S₅₂ = 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, … 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) – 1/2
Solution:

Or a²ⁿ⁺² + b²ⁿ⁺² + 2aⁿ⁺¹ bⁿ⁺¹
= ab [a²ⁿ + b²ⁿ + 2aⁿbⁿ]
Or a²ⁿ a² + b²ⁿ b² + 2ab.aⁿbⁿ = ab[a²ⁿ + b²ⁿ + 2aⁿbⁿ]
Or a²ⁿa² – ab.a²ⁿ + b²ⁿb² – ab.b²ⁿ
+ 2ab.aⁿbⁿ – 2ab.aⁿbⁿ = 0
Or a²ⁿ a(a-b) + b²ⁿ b(b-a) = 0
⇒ a²ⁿ⁺¹ (a-b) – b²ⁿ⁺¹ (a-b) = 0
Or (a-b) [a²ⁿ⁺¹ – b²ⁿ⁺¹] = 0
Or a²ⁿ⁺¹ = b²ⁿ⁺¹
[if a ≠ b ⇒ a – b ≠ 0]
Or a2n+1/b2n+1 = 1

Comparing powers of both sides
2n + 1 = 0
Or n = – 1/2
Thus, value of n = – 1/2
Hence option (d) is correct

Question 17.
If G₁ and G₂ are two GM between a and b then value of G₁G₂ is :
(a) √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 1/2,5/13,5/16, …. is
(a) A.P.
(b) GP.
(c) H.P.
(d) Other
Solution:

Question 20.
6th term of series 1,1/4,1/7,1/10,….. is:
(a) 1/13
(b) 1/16
(c) 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 a₆ = a + 5d
= 1 + 5 × 3 = 1 + 15 = 16
Thus, 6th term of corresponding H.P. is 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 + G² = 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 = √AH
G² = 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 H/a + H/b is :
(a) 2
(b) a+b/ab
(c) 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 = 2ac/a+c ….(i)
G.M. of a,b,c
G = √ac ….(ii)
From equation (i) and (ii),
b = 2ac/a+c
We know that GM. > H.M.
G > H
√ac> b
Hence, option (c) is correct

Question 27.
If nth term of any series is n2/3n 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 m² : n², 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 + a²
(∵ hypotenuse is longest side)

Question 31.
– 2/7, a, – 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 3².
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, 1/2, 1/8, 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 G₁ and G₂ are two geometric mean between a and b, then prove that G₁,G₂ = 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.
A₁,A₂, are two A.M. between two positive numbers a and b two GM. G₁, G₂ and two H.M. H₁, H₂, then prove that :
A₁H₂ = A₂H₁ = G₁G₂ = ab
Solution:
Let two numbers are a and b, then
A₁, A₂ are two A.M. between a and b, then

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