Solution

SEM 1 ( SETS, RELATION&FUNCTION, TRIGONOMETRY, COMPLEX NUMBER, INEQUALITY, SEQUENCE

AND SERIES, P&C)

Class 11 - Mathematics
Section A
1.
(c) 2n
Explanation: The total no of subsets = 2n
2.
(b) {3, 6, 9, 12, 18, 21, 24, 27}

OR
Explanation: Since set B represent multiple of 5 so from Set A common multiple of 3 and 5 are excluded.
3.
(d) domain (R) = {1, 3, 4}
Explanation: The set of all first elements of the ordered pairs in R is called domain of relation.

AT
i.e. domain (R) : {a : (a, b) ∈ R}
∴ Domain (R) = {1, 3, 4}

4.

5.
(c) 8
UC
Explanation: Since A has 3 elements and B has 2 elemets, then number of functions from A to B is 2 3
= 8

(b) 3
ED
x+1
Explanation: f (x) = x−1
x +1
+1
f (x)+1 x −1
f (f (x)) = =
f (x)−1 x +1
−1
x −1

x+1+x−1 2x
= = = x
x+1−x+1 2

x+1
AD

f (f (f (x))) = f (x) =
x−1

2+1
f (f (f (2))) =
2−1

=3
6.
(c) sin (α − β) = 0
AS

Explanation: Given sin α = sin β and cos α = cos β

Now sin α ⋅ cos β = cos α ⋅ sin β
⇒ sin α ⋅ cos β − cos α ⋅ sin β = 0

⇒ sin(α − β) = 0
PR

7.
(c) 1

Explanation: 2 sin 5π

12
sin
π

12

5π 5π
= {cos(
12
−
π

12
) − cos(
12
+
π

12
)} [since 2sin Asin B= cos (A - B) - cos (A + B)]
π π 1 1
= (cos − cos )= ( − 0) =
3 2 2 2

√3
8. (a) 2

Explanation: Using (3 sin θ − 4 sin 3

θ) = sin 3θ , we get
(3 sin 40° - 4 sin340°) = sin(3 × 40 )
∘
= sin 120°
√3
= sin (180° - 60°) = sin 60° = 2

9.
(b) arg (z1) = arg (z2)

1/9
 Explanation: Let z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2)
Since |z1 + z2| = |z1| + |z2|
⇒ z1 + z2 = r1 cos θ1 + ir1 sin θ1+ r2 cos θ2 + ir2 sin θ2
⇒ |z1 + z2|
−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−
= √r 2
1
cos
2
θ1 + r
2
2
cos
2
θ2 + 2r1 r2 cos θ1 cos θ2 + r
2
1
sin
2
θ1 + r
2
2
sin
2
θ2 + 2r1 r2 sin θ1 sin θ2

−−−−−−−−−−−−−−−−−−−−−
= √r 2
1
+ r
2
2
+ 2r1 r2 cos(θ1 − θ2 )

But |z1 + z2| = |z1| + |z2|

−−−−−−−−−−−−−−−−−−−−−
2 2
⇒ √r + r + 2r1 r2 cos(θ1 − θ2 ) = r1 + r2
1 2

Squaring both sides,

2 2 2 2
⇒ r + r + 2r1 r2 cos(θ1 − θ2 ) = r + r + 2r1 r2
1 2 1 2

⇒ 2r1r2 – 2r1r2 cos (θ1 – θ2) = 0

OR
⇒ 1 – cos (θ1 – θ2) = 0
⇒ cos (θ1 – θ2) = 1
⇒ (θ1 – θ2) = 0
= θ2

AT
⇒ θ1

∴ arg (z1) = arg (z2)

10.
(b) (
−3

25
+

Explanation: (1 - 2i)-2 =
4

25
i)

(1−2i)
2
= 1

2
(1+4i −4i)
=
UC 1

(1−4−4i)
= 1

(−3−4i)
×
(−3+4i)

(−3+4i)

(−3+4i) (−3+4i) −3 4
= 2
= 25
=( 25
+
25
i)
(9−16i )
ED
11.
−2
(b) ( 3
, ∞)

Explanation: (x + 3) + 4 > -2x + 5

⇒ x + 7 > -2x + 5

⇒ x + 7 + 2x > -2x + 5 + 2x
AD

⇒ 3x + 7 > 5

⇒ 3x + 7 - 7 > 5 - 7
⇒ 3x > -2
−2
⇒ x >
AS

−2
⇒ x ∈ ( , ∞)
3

12. (a) − 1

6
≤ x <
1

Explanation: -2 ≤ 6x - 1 < 2
PR

⇒ -2 + 1 ≤ 6x - 1 + 1 < 2 + 1

⇒ -1 ≤ 6x < 3
−1 6x 3
⇒ ≤ <
6 6 6
−1 1
⇒ ≤ x <
6 2

13.
(b) 3
Explanation: 2730 = (15 × 14 × 13) = 15pr
⇒ r = 3.
14.
(b) 69,760
Explanation: As we have the ten digits from 0 to 9 which can be used
Number of five digit telephone numbers in which the digits can be repeated = 105 = 100000
Number of five digit telephone numbers in which the digits cannot be repeated = 10P5 = 10!
= 6 × 7 × 8 × 9 × 10 =
(10−5)!

2/9
 30240
Therefore the number of five digit telephone numbers in which at least one digit is repeated = 100000 - 30240 = 69760
15.
(d) 5
Explanation: If you join every distinct pair of vertices you will get n
C2 lines.
These n
C2 lines account for the n sides of the polygon as well as for the diagonals.
n(n−1) n(n−3)
So the number of diagonals is given by n
C2 − n =
2
− n=
2

But given number of sides = number of diagonals =n

n(n−3)
⇒ n=
2

⇒ 2n = n(n − 3)

⇒ n− 3 = 2

⇒ n= 5

OR
16.
(c) a rational number
Explanation: We have (a + b)n + (a - b)n

AT
=[ C a + C a b+ C a b +
n
0
n n
1
n−1 n
2
n−2 2 n
C3 a
n−3
b
3
+ ..... + n
Cn b ]
n
+
[
n
C0 a
n
−
n
C1 a
n−1
b +
n
C2 a
n−2
b
2
−
n
C3 a
n−3 3
b + ..... + (−1) n
⋅
n
Cn b
n
]
= 2[ n
C0 a
n
+
n
C2 a
n−2 2
b + ...]
–
Let a = √5 and b = 1 and n = 4
–

= 2[25 + 30 + 1] = 112
–
Now we get (√5 + 1) + (√5 − 1) = 2 [ 4 4 4
UC– 4 4
– 2 2 4
– 0 4
C0 (√5) + C2 (√5) 1 + C4 (√5) 1 ]

17. (a) x < y

Explanation: Given x = 9950 + 10050 and y = (101)50
ED
Now y = (101)50 = (100 + 1)50 = 50
C0 (100)
50
+
50
C1 (100)
49
+ 50
C2 (100)
48
+ .... + 50
C50 .....(i)
Also (99)50 = (100 - 1)50 = = 50
C0 (100)
50
−
50
C1 (100)
49
+ 50
C2 (100)
48
- .... + 50
C50 ....(ii)
Now subtract equation (ii) from equation (i), we get
(101)50 - (99)50 = 2 [ 50
C1 (100)
49
+
50
C3 (100)
47
+ …]
AD

= 2 [50(100) 49
+
50×49×48

3×2×1
(100)
47
+ …]

= (100) 50
+ 2(
50×49×48

3×2×1
(100)
47
)

⇒ (101)50 - (99)50 > (100)50

AS

⇒ (101)50 > (100)50 + (99)50

⇒ y>x
18.
(c) 5
PR

Explanation: Let (a + ar + ar2 + ... ∞ ) = 15 and (a2 + a2r2 + a2r4 + .... .∞ ) = 45.
2

Then, a
= 15 and a

2
= 45 .
(1−r) (1− r )

2 (1−r)
On dividing, we get a

2
×
a
=
45

15
⇒
a
= 3
(1− r ) (1+r)

15(1−r)
⇒ = 3 [using a
= 15 ]
(1+r) (1−r)

⇒ 3 + 3r = 15 - 15r ⇒ 18r = 12 ⇒ r= 2

∴
a

2
= 15 ⇒ 3a = 15 ⇒ a = 5 .
(1− )
3

Therefore, the required first term is 5.

19.
(d) 512
– –
Explanation: Given GP is 2, 2√2, 4, 4√2, ...
2√2 –
Here, a = 2 and = 2
= √2

3/9
 – 16
∴ T17 = ar
16
= 2 × (√2)
8
= 2 × 2 = 2
9
= 512
Therefore, the required 17th term is 512.
20.
1
(b) 400
1
Explanation: By inspection we get an = 2
, as
n2 ⋅(n+1)

a1 = 2
1

2
=
1

4
1 ⋅(1+1)

a2 = 2
1

2
=
1

36
2 ⋅(2+1)

and a3 = 2
1

2
=
144
1

3 ⋅(3+1)

1 1
∴ a4 = 2 2
=
400
4 ⋅(4+1)

Section B
21. According to the question,

OR
A = {1, 2, 3, 4, 5} and B = {2, 4,6}
Now, A-B represents a set of only those elements of A which does not belong to B.

∴ A - B = {1, 2, 3, 4, 5} - {2, 4, 6}

AT
⇒A - B = {1, 3, 5}
whereas, B-A represents set of only those elements of B which do not belong to A.
B - A = {2, 4, 6} - {1, 2, 3, 4, 5} = {6}
22. From the given we can assume,
Let x be a pre-image of 6 Then
f(x) = 6 = x2 - 2x - 3 = 6 = x2 - 2x - 9 = 0 = x = 1 ±√10
−−
−−
UC
Since x = 1 ± √10 ∉ A so there is nor pre image of 6
f(x) = -3 = x2 - 2x - 3 = -3 = x2 - 2x = 0 = x = 0.2
ED
Clearly, 0.2 ∈ A So 0 and 2 are pre image of -3
Let x be a pre image of 5 then
f(x) = 5 = x2 - 2x - 3 = 5 = x2 - 2x - 8 = 0 = (x - 4) (x + 2) = 0 = x = 4,
Since, -2A be 4A so, -2 is a pre image of 5
AD

23. To prove: 4 sin sin + 3 cos tan

π

6
+ cosec
2 π

3
π

3
π

4
2 π

2
= 4

π π π π π
Now, L.H.S = 4 sin 6
sin
2

3
+ 3 cos
3
tan
3
+ cosec
2

Put π = 180 and converting into degree,

∘

∘ ∘ ∘ ∘ ∘
180 2 180 180 180 2 180
= 4 sin sin + 3 cos tan + cosec
6 3 3 3 2
AS

= 4 sin 30° sin2 60° + 3 cos60° tan 45° + cosec2 90°

2
1 √3 1 2
= 4 × × ( ) +3 × × 1 + (1)
2 2 2

3+3+2
= = 2
2

LHS = RHS
PR

24. We have, 4x - 1 ≤ 0
⇒ 4x ≤ 1
⇒ x ≤ ⇒ x ∈(-∞ ,
1

4
1

4
] ...... (i)
Also, 3 - 4x < 0
⇒ 0 > 3 - 4x

⇒ 4x > 3

⇒x >
3

⇒ x ∈ ( , ∞ ) ......(ii)
3

Hence, the solution set of inequalities is the intersection of (i) and (ii). But, (- ∞ , 1

4
) ∩ ( , ∞) = ϕ
3

Thus, the given set of inequations has no solution.

25. Substituting x = 1 and -1 in (1 - x + x2)n = a0 + a1x + a2x2 + ... + a2n x2n we obtain
1 = a0 + a1 + a2 + ... + a2n ... (i)
and

4/9
 3n = a0 - a1 + a2 - ... + a2n ... (ii)
Adding (i) and (ii), we obtain
3n + 1 = 2 (a0 + a2 + ... + a2n)
n
3 +1
Therefore, the value of a0 + a2 + a4 + ... + a2n is 2

Section C
26. The Venn diagram for (A∪B) The shaded portion represents (A∪B)'

OR
27. Here we have, f(x) = 0 for all x ∈ R
Then, dom (f) = R and range (f) = {0}
Now, we have,

AT
X -2 -1 0 1 2

f(x) = 2 0 0 0 0 0
On a graph paper, we draw the horizontal line X' OX as the x-axis and the vertical line YOY' as the y-axis.
Taking the scale: 10 small divisions = 1 unit.
UC
Now, on this graph paper, we plot the points A(-2, 0), B(-1, 0), O(0, 0), C(1, 0) and D(2,0) and join them successively to get the
graph line ABOCD, shown below, whose equation is y = 0.
ED
AD

Graph of the function, f(x) = 0

AS

28. We have L.H.S. = cot 4x (sin 5x + sin 3x)

cos 4x 5x+3x 5x−3x
= [2 sin( ) cos( )]
sin 4x 2 2

cos 4x
= [2 sin 4x cos x] = 2 cos 4x cos x
PR

sin 4x

We have R.H.S. = cot x [sin 5x - sin 3x]

cos x 5x+3x 5x−3x
= [2 cos( ) sin( )]
sin x 2 2

C+D C−D
[∵ sin C − sin D = 2 cos ⋅ sin ]
2 2

=
cos x

sin x
[2 cos 4x sin x] = 2 cos 4x cos x
Hence L.H.S. = R.H.S
(5x−2) (7x−3)
29. Here x

2
⩾
3
−
5
x 5x 2 7x 3
⇒ ⩾ − − +
2 3 3 5 5
15x−50x+42x −10+9
⇒ ⩾
30 15

7x −1
⇒ ⩾
30 15

Multiplying both sides by 30, we have

7x ⩾ −2

Dividing both sides by 7, we have

The solution set is [ −2

7
, ∞)

5/9
 The representation of the solution set on the number line is

30. Total letters in the word PERMUTATIONS = 12.

Here T = 2
(i) Now first letter is P and last letter is S which are fixed.
So the remaining 10 letters are to be arranged between P and S.
∴ Number of Permutations

=
10!
=
10×9×8×7×6×5×4×3×2!
= 1814400
2! 2!

(ii) There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.
∴ Number of permutations of all vowels together = P
5
5

5!
= = 5 × 4 × 3 × 2 × 1 = 120
0!

Now consider the 5 vowels together as one letter. So the number of letters in the word when all vowels are together = 8.
∴ Number of Permutations =
8! 8×7×6×5×4×3×2!

OR
= = 20160
2! 2!

Hence the total number of permutations = 120 × 20160 = 2419200

(iii) Here P and S are on 1 st and 6 th places P and S are on 2 nd and 7 th places
P and S are on 3 rd and 8 th places P and S are on 4 th and 9 th places

AT
P and S are on 5 th and 10 th places P and S are on 6 th and 11 th places
P and S are on 7 th and 12 th Places
Now we see that P and S can be put in 7 ways and also P and S can interchange their positions.
∴ Number of permutations = 2 × 7 = 14

Now the remaining 10 places can be filled with remaining 10 letters.

∴ Number of permutations = =
10!

2!
= 1814400
10×9×8×7×6×5×4×3×2!

2!
UC
Thus total number of permutations = 14 × 1814400 = 25401600
31. Here, it is given: Three numbers are in A.P. Their sum is 21.
ED
Suppose the numbers be a - d, a, a + d
According to the question
a + d + a + a - d = 21
⇒ 3a = 21 ⇒ a = 7
Therefore numbers are 7 - d, 7, 7 + d
AD

As per question when second number is reduced by 1 and third is increased by 1 then the numbers become-
7 - d, 7 - 1, 7 + d + 1
⇒ 7 - d, 6, 8 + d

These above numbers are in GP

AS

Therefore, 62 = (7 - d)(8 - d) (When a, b, c are in GP, b2 = ac)

⇒ 36 = 56 + 7d - 8d - d2
⇒ d2 + d - 20 = 0 ⇒ d2 + 5d - 4d - 20 = 0
PR

⇒ d(d + 5) - 4(d + 5) = 0 ⇒ (d - 4) (d + 5) = 0

⇒ d = 4, or d = -5

When taking d = 4, then we get numbers are

7 - d, 7, 7 = d = 7 - 4, 7, 7 + 4 = 3, 7, 11
When taking d = -5, the we get numbers are
7 - d, 7, 7 + d = 7 - (-5), 7, 7 + (-5) = 12, 7, 2
Therefore, the two sets of triplet are 3, 7, 11 and 12, 7, 2
Section D
32. i. Given that, sin(A + B) = 1
π
⇒ sin(A + B) = sin 2

⇒ A+B= π

2
...(i)
and sin(A - B) = 1

⇒ sin(A - B) = sin π

6
π
⇒ A-B= 6
...(ii)

6/9
 On adding Eqs. (i) and (ii), we get
2π π
2A = ⇒ A=
3 3

ii. tan(A + 2B)⋅ tan (2A + B)

π π 2π π
= tan( + ) tan( + )
3 3 3 6

2π 5π π π π π
= tan( ) tan( ) = tan( + ) tan( + )
3 6 2 6 2 3

π π –
= (− cot
6
) (− cot
3
) = (− √3) (−
1
) =1
√3

iii. sin2A - sin2B

2 π 2 π
= sin ( ) − sin ( )
3 6

2
√3 1 2 3 1 2 1
= ( ) − ( ) = − = =
2 2 4 4 4 2

OR
cos 2 A = 2 cos2 A - 1

OR
2 π
= 2 cos ( ) − 1
3

2
1 2 1 1
2( ) − 1 = − 1 = − 1 = −
2 4 2 2

33. i. i30 = (i)4× 7i2 = -1

AT
ii. i−39 = i(i−40)
= i((i2)−20) = i((−1)−20) [∵ i2 − 1]
= i( ) = i( ) = i = 0 + i(1)
1

20
1

1
(−1)

Comparing with a + ib,

a = 0, b = 1
0+i
UC
iii. z = i9 + i19,
ED
= (i4)2.i + (i4)4.i3
= i + i3 [∵ i4 = 1]
= i − i [∵ i3 = −i]
=0
AD

= 0 + 0i
OR
Remember that i2 = −1 hence
i37 = (i36)⋅i = (i2)18.i = (−1)18⋅i = i
AS

34. i. Let x litres of 4% boric acid solution is required to be added.

Then, total mixture = (x + 750) litres
This resulting mixture is to be more than 5% but less than 8% boric acid.
Total amount of acid = 750 of 10% + x of 4%
PR

ATQ
5 750×10+4x 8

100(750+x)
< 100
< 100(750+x)

3750 + 5x < 7500 + 4x; 7500 + 4x < 6000 + 8x

x < 3750; x > 375
375 and 3750
ii. Given: C = (F − 32) 5

Given: The solution is to be kept between 30o C and 35o C.

i.e., 30o < C < 35o
5 ∘
⇒ 30 < (F − 35) < 35
9

Multiplying 9

5
on all sides.
9 9 5 9
⇒ 30 × < × (F − 32) < 35 ×
5 5 9 5

⇒ 9 × 6 < F - 32 < 7 × 9
⇒ 54 < F - 32 < 63

⇒ 54 < F -32 < 63

7/9
 Adding 32 on all sides
⇒ 54 + 32 < F -32 + 32 < 63 + 32

⇒ 86 < F < 95

∴ The required range of temperature is between 86o F and 95o F.

iii. −8x + 3 ≥ 27 and −13x + 5 ≥ 57
x ≤ −4

OR
OR

AT
Section E
35. Let z = x + iy
Given: |z| = z + 1 + 2i
⇒ |x + iy| = x + iy + 1 + 2i
−−−−−−
UC
2
⇒ √x + y
2
= (x + 1) + i(y +2)
⇒ x2 + y2 = (x + 1)2 + 2i (x + 1) (y + 2) − (y + 2)2 [Squaring both sides]
ED
⇒ x2 + y2 = x2 + 2x + 1 + 2i (xy + 2x + y + 2) − (y2 + 4y + 4)
⇒ 2y2 − 2x + 4y + 3 = 2i (xy + 2x + y + 2)
⇒ y2 − x + 2y + 2 = i(xy + 2x + y + 2)
⇒ (y2 − x + 2y + 2) − i(xy + 2x + y + 2) = 0
AD

On comparing we get,
(xy + 2x + y + 2) = 0
⇒ (x + 1) (y + 2) = 0
⇒ x = − 1, y = −2
AS

Also, (y2 − x + 2y + 2) = 0
Taking x = - 1, (y2 - (- 1) + 2y + 2) = 0
⇒ (y2 + 2y + 3) = 0
Does not have a solution since roots will be imaginary.
PR

Taking y = −2, (4 − x − 4 + 2 ) = 0
⇒ x = 2

∴ z = x + iy = 2 - 2i

36. i. Given that, the total number of player is 17.

We have to select 11 players including exactly 4 bowlers.
Hence, 4 bowlers will be selected from 5 bowlers and the remaining 7 players will be selected from 12 batsmen.
Now, 4 bowlers out of 5 bowlers can be selected in 5C4 ways.

7 players out of 12 players can be selected in 12C7 ways.

∴ Total number of ways selecting the 11 players
=
5
C4 ×
12
C7 =
5
C1 ×
12
C5 [∵ nCr = nCn - r]
12! 12×11×10×9×8
= 5 × = 5 ×
5!7! 5×4×3×2×1

8/9
 = 5 × 11 × 9 × 8 = 55 × 72 = 3960 ways
Hence, he can select a team of 11 players in 3960 ways.
ii. A group of people who works together with energy and passion together with energy and passion to achieve their goals form a
team spirit.
– – 7
37. To find: Value of (2 + √3) 7
+ (2 − √3)

Formula used: n
Cr =
(n−r)!(r)!
n!

(a + b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn

(a + b)7 = [ 7 7
C0 a ] + [ C1 a
7 7−1
b] + [ C2 a
7 7−2 2
b ]+ [ C3 a
7 7−3 3
b ] + [ C4 a
7 7−4
b ]
4

7 7−5 5 7 7−6 6 7 7
+ [ C5 a b ] + [ C6 a b ] + [ C7 b ]

= 7C0a7 + 7C1a6b + 7C2a5b2 + 7C3a4b3 + 7C4a3b4 + 7C5a2b5 + 7C6a1b6 + 7C7b7 … (i)

(a - b)7 [ 7 7
C0 a ] + [ C1 a
7 7−1
(−b)] + [ C2 a
7 7−2 2
(−b ) ] + [ C3 a
7 7−3 3
(−b ) ] + [ C4 a
7 7−4 4 7 7−5
(−b) ] + [ C5 a
5
(−b) ]

7 7−6 6 7 7

OR
+ [ C6 a (−b ) ] + [ C7 (−b ) ]

= 7C0a7 - 7C1a6b + 7C2a5b2 - 7C3a4b3 + 7C4a3b4 - 7C5a2b5 + 7C6a1b6 - 7C7b7 ….. (ii)
Adding eqn. (i) and (ii)
(a+b)7 + (a-b)7= [7C0a7 + 7C1a6b + 7C2a5b2 + 7C3a4b3 + 7C4a3b4 + 7C5a2b5 + 7C6a1b6 + 7C7b7] + [7C0a7 - 7C1a6b + 7C2a5b2 -

AT
7C a4b3 + 7C a3b4 - 7C a2b5 + 7C a1b6 - 7C b7]
3 4 5 6 7

= 2[7C0a7 + 7C2a5b2 + 7C4a3b4 + 7C6a1b6]

2
7! 7 7! 5b 7! 3 4 7! 1 6
= 2 [[ a ] + [ a ]+[ a b ] + [ a b ]]
0!(7−0)!

= 2[(1)a7 + (21)a5b2 + (35)a3b4 + (7)ab6]

= 2[a7 + 21a5b2 + 35a3b4 + 7ab6] = (a+b)7 + (a-b)7
2!(7−2)! 4!(7−4)!

UC 6!(7−6)!

–
Putting the value of a = 2 and b = √3 in the above equation
– 7 – 7
ED
(2 + √3) + (2 − √3)
7 5
– 2 3
– 4 – 6
= 2 [{2 } + {21(2) (√3) } + {35(2) (√3) } + {7(2)(√3) }]

= 2[128 + 21(32)(3) + 35(8)(9) + 7(2)(27)]

= 2[128 + 2016 + 2520 + 378]
= 10084
AD

38. i. Sn = 5 + 55 + 555 + ………. up to n terms

= 5 [1 + 11 + 111 + ………. up to n terms]
=
5

9
[9 + 99 + 999 + ....... up ton terms]
=
5
[(10 - 1) + (102 - 1) + (103 - 1) + ...... up to n terms]
AS

9
5 n
10( 10 −1)
= 9 [ − n]
10−1

5 10 n
= [ (10 − 1) − n]
9 9

50 n 5
= (10 − 1) − n
PR

81 9

ii. Sn = .6 + .66 + .666 + …………. up to n terms

= 6 [.1 + .11 + .111 + ………. up to n terms]
=
6

9
[.9 + .99 + .999 + ....... up to n terms]
6 9 99 999
=
9
[
10
+
100
+
1000
+ ……… up to n terms ]
6 1 1 1
=
9
[(1 −
10
) + (1 −
2
) + (1 −
3
)…… up to n terms ]
10 10

=
6

9
[n − (
10
1
+
1

2
+
1

3
+ ……… up to n terms )]
10 10

1 1
(1− )
2 10 10n
= [n − ]
3 1
1−
10

2 1 1
= [n − (1 − )]
3 9 2
10

2n 2 1
= − (1 − n
)
3 27 10

9/9