RBSE Solutions for Class 11 Maths Chapter 8 Sequence, Progression, and Series Ex 8.2

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 =  \(\frac { n }{ 2 }\)
S₂₀ = \(\frac { 20 }{ 2 }\)
= 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) = \(\frac { 96 }{ 3 }\)
=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 = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
⇒ \(\frac { n }{ 2 }\) [2 × 1 + (n – 1) × 5] = 148
⇒ \(\frac { n }{ 2 }\) [2 + 5n – 5] = 148
⇒ n (5n – 3) = 296
⇒ 5n² – 3n – 296 = 0
On solving,

Taking ‘+’, n = 8
Taking ‘-‘, n = –\(\frac { 37 }{ 5 }\)
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= \(\frac { mn }{ 2 }\) (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 =  \(\frac { 216 }{ 24 }\)
⇒ 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 = \(\frac { x + z }{ 2 }\)
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 = \(\frac { 30 }{ 2 }\) (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 = \(\frac { n }{ 2 }\) [2a + (n – 1)d]
⇒ (n – 2) 360° = \(\frac { n }{ 2 }\) [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.

RBSE Solutions for Class 11 Maths