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# 10th Maths July Month Assignment 2

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- Class: 10
- Subject : Maths
- Unit: 2 Numbers and Sequence

### UNIT 2 : NUMBERS AND SEQUENCE

__Part - A__

####
__I. One Mark Questions__

###
**1.The sum of the exponents of the prime factors in the prime
factorization of 1729 is**
a) 1
b)2
c)3
d)4

**2.If the number 4n isend with the digit 6,n****Є***N***, then the value of n is**
a)n is even
b)n is odd
c)n is prime
d)No value

**3.In the sequence In the sequence Fn = Fn-1 + Fn-2,ifF1=1, F2=3, then F5 is**
a) 3
b) 5
c) 8
d) 11

**4.If an A.P is called constant Arithmetic Progression, then the
common difference is**
a) 5
b) 2
c) 1
d) 0

**5.In an AP, the first term is 20 and the common difference is 8.
Then the 4th term is**
a) 20
b) 28
c) 44
d)36

**6.In an A.P, if every term is added by the same constant then the
resulting sequenceforms,**
a) G.P
b) AP
c) AP and GP
d) None

**7.In the Geometric Progression a, ar, ar2 …arn-1 then r iscalled as**
a) First term
b) Common difference
c) Common ratio
d) Last term

**8.The next term of the sequence 3/16, 1/8, 1/16, 1/18 is**
a) 1/24
b) 1/27
c) 2/3
d) 1/81

**9. The value of ****(1**^{3}+2^{3}+3^{3} …+15^{3}**) – (1+2+3 …+15) is**
a) 14400
b) 14200
c) 14280
d) 14520

**10. The sum of
3+1+1/3+…+is**
a) 0
b) 1/3
c) 9/2
d) 2/9

**1.The sum of the exponents of the prime factors in the prime factorization of 1729 is**

a) 1

b)2

c)3

d)4

**2.If the number 4n isend with the digit 6,n**

**Є**

*N***, then the value of n is**

a)n is even

b)n is odd

c)n is prime

d)No value

**3.In the sequence In the sequence Fn = Fn-1 + Fn-2,ifF1=1, F2=3, then F5 is**

a) 3

b) 5

c) 8

d) 11

**4.If an A.P is called constant Arithmetic Progression, then the common difference is**

a) 5

b) 2

c) 1

d) 0

**5.In an AP, the first term is 20 and the common difference is 8. Then the 4th term is**

a) 20

b) 28

c) 44

d)36

**6.In an A.P, if every term is added by the same constant then the resulting sequenceforms,**

a) G.P

b) AP

c) AP and GP

d) None

**7.In the Geometric Progression a, ar, ar2 …arn-1 then r iscalled as**

a) First term

b) Common difference

c) Common ratio

d) Last term

**8.The next term of the sequence 3/16, 1/8, 1/16, 1/18 is**

a) 1/24

b) 1/27

c) 2/3

d) 1/81

**9. The value of**

**(1**

^{3}+2^{3}+3^{3}…+15^{3}**) – (1+2+3 …+15) is**

a) 14400

b) 14200

c) 14280

d) 14520

**10. The sum of 3+1+1/3+…+is**

a) 0

b) 1/3

c) 9/2

d) 2/9

####
__Part – B__

####
__II. Short Answer.__

#### 11. Find the Highest Common Factor (HCF) of 340 and 412 by using Euclid’s Division Algorithm.

###
We get 412 = 340 × 1 + 72

The remainder 72 ≠ 0

Again applying Euclid’s division algorithm

340 = 72 × 4 + 52

The remainder 52 ≠ 0.

Again applying Euclid’s division algorithm

72 = 52 × 1 + 20

The remainder 20 ≠ 0.

Again applying Euclid’s division algorithm,

52 = 20 × 2 + 12

The remainder 12 ≠ 0.

Again applying Euclid’s division algorithm.

20 = 12 × 1 + 8

The remainder 8 ≠ 0.

Again applying Euclid’s division algorithm

12 = 8 × 1 + 4

The remainder 4 ≠ 0.

Again applying Euclid’s division algorithm

8 = 4 × 2 + 0

The remainder is zero.

Therefore H.C.F. of 340 and 412 is 4.

12. If 13824 = 2a x 3b, then find
the value of a and b.
Using factor tree method factorisee 13824

13. Solve 8x = 1 ( mod 11)

15. Which term of an A.P 16, 11,
6, 1…is -54?
__Solution:__
A.P = 16, 11,6, 1, ………..
It is given that
tn = -54
a = 16,
d = t2 – t1
= 11 – 16
d = -5
∴ tn = a + (n – 1)d
-54 = 16 + (n – 1) (-5)
-54 = 16 – 5n + 5
21 – 5n = -54
-5n = -54 -21
-5n = -75
n = \(\frac { 75 }{ 5 } \) =15
∴ 15th term is -54.

16. Find the sum of first six
terms of GP 5, 15, 45 …

17. If 1+2+3+…+n= 666 then find n.

The remainder 72 ≠ 0

Again applying Euclid’s division algorithm

340 = 72 × 4 + 52

The remainder 52 ≠ 0.

Again applying Euclid’s division algorithm

72 = 52 × 1 + 20

The remainder 20 ≠ 0.

Again applying Euclid’s division algorithm,

52 = 20 × 2 + 12

The remainder 12 ≠ 0.

Again applying Euclid’s division algorithm.

20 = 12 × 1 + 8

The remainder 8 ≠ 0.

Again applying Euclid’s division algorithm

12 = 8 × 1 + 4

The remainder 4 ≠ 0.

Again applying Euclid’s division algorithm

8 = 4 × 2 + 0

The remainder is zero.

Therefore H.C.F. of 340 and 412 is 4.

__Solution:__

A.P = 16, 11,6, 1, ………..

It is given that

tn = -54

a = 16,

d = t2 – t1

= 11 – 16

d = -5

∴ tn = a + (n – 1)d

-54 = 16 + (n – 1) (-5)

-54 = 16 – 5n + 5

21 – 5n = -54

-5n = -54 -21

-5n = -75

n = \(\frac { 75 }{ 5 } \) =15

∴ 15th term is -54.

16. Find the sum of first six
terms of GP 5, 15, 45 …

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