## 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} = \(\frac { n }{ 2 }\)

S_{20} = \(\frac { 20 }{ 2 }\)

= 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, T_{1} =2 × 1 +3 = 2 + 3 = 5

and T_{2 }= 2 × 2 + 3 = 4 + 3 =7

Common difference

d = T_{2 }– T_{1} = 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 n^{2} + 2n. Find the first term and common difference.

Solution:

Sum of terms

S_{n}= n^{2} + 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}= (n^{2 }+ 2n) – [(n – 1)^{2} + 2(n − 1)]

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

= [n^{2} + 2n] – [n^{2} – 1]

= n^{2} + 2n – n^{2} + 1

= 2n + 1

First term T_{1}= 2 × 1 + 1 = 2 + 1 = 3

Second term, T_{2} = 2 2 + 1 = 4 + 1 = 5

Common difference d = T_{2 }– T_{1} = 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

⇒ 5n^{2} – 3n – 296 = 0

On solving,

Taking ‘+’, n = 8

Taking ‘-‘, n = –\(\frac { 37 }{ 5 }\)

Negative value of n is impossible.

Hence, n = 8

Last term= T_{9} = 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 a_{1} 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 S_{1}, S_{2} and S_{3} respectively, then prove that S_{3} = 3 (S_{2 }– S_{1}).

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

= S_{3 }[From equation (iii)]

Hence, 3 (S_{2} – S_{1}) = S3

Hence Proved.

Question 8.

If sum of m A.P. of n terms are S_{1}, S_{2} and S_{3}…, S_{m} respectively. Their first term are 1, 2, 3,…., m respectively and common difference 1, 3, 5,…., (2m – 1) respectively, then prove that

S_{1} + S_{2} + S_{3} + … + 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} + a^{3} + (a + d)^{3} = 408

⇒ (4 – d)^{3} + (4^{3}) + (4 + d)^{3} = 408

⇒ (4^{3} – 3(4^{2})d + 3 × 4 × d^{2 }– d^{3}) + (4^{3})

+ (4^{3} + 3(4^{2})d + 3 × 4 × d^{2} + d^{3}) = 408

⇒ 3(4^{3}) + 24d^{2 }= 408

⇒ 24d^{2} = 408 – 192 = 216

⇒ d^{2 }= \(\frac { 216 }{ 24 }\)

⇒ d^{2 }= 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, A_{1} + A_{2} + A_{3} + …, 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 x^{2} (y + z), y^{2} (z + x), z^{2}(x + y) are in A.P., then prove that either x, y, z are in A.P. or xy + yz + zx = 0.

Solution:

∵ x^{2}(y + z), y^{2}(z + x), z^{2}(x + y) are in A.P.

∴ Adding xyz in each terms x^{2}(y + z) + xyz, y^{2}(z + x) + xyz, z^{2}(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. a_{1} + a_{2} + a_{3 }…, A_{30}.

Given that

a_{1} + a_{7} + a_{10 + }a_{21} + a_{24} + 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.,

^{T}r ^{+Tr – 1} = a + 1

∵ a_{7} is 7th term from start and a_{24} is 7th term from last

a_{7} + a_{24} = a_{1 }+ a_{30} …(i)

Similarly, a_{10} is 10th term from start and a_{21} is 10th term from last, then

a_{10} + a_{21} = a_{1 }+ a_{30}

a_{1} + a_{7} + a_{10} + a_{21} + a_{24} + a_{30} = 540

⇒ (a_{1}+ a_{30}) + (a_{7 }+ a_{24}) + (a_{10 }+ a_{21}) = 540^{
}From equation (i) and (ii),

(a_{1 }+ a_{30}) + (a_{1 }+ a_{30}) + (a_{1}+ a_{30}) = 540

⇒ (a_{1 }+ a_{30}) = 540

⇒ a_{1 }+ a_{30} = 180

Hence, sum of 30 terms

S_{30 }= \(\frac { 30 }{ 2 }\) (a_{1 }+ a_{30})

= 15 (180) = 2700

Hence, a_{1} + a_{2} + a_{3} + …….+ a_{30} = 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°

S_{1} = \(\frac { n }{ 2 }\) [2a + (n – 1)d]

⇒ (n – 2) 360° = \(\frac { n }{ 2 }\) [2 x 52° + (n – 1) 8°]

⇒ 360°n – 720° = 104° n + 8°n^{2} – 8°n

⇒ 8°n^{2} + (104° – 360° – 8°)n + 720° = 0

⇒ 8°n^{2} – 264°n + 720° = 0

⇒ n^{2} – 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, number of sides is 3.

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