10th Maths July Month Assignment Answer key

10th Maths July Month Assignment 2

Answer key 

10th Maths July Month Assignment 2 Answer key
10th July Month Assignment Answer Key

10th Standard Maths July Month Assignment 2 Question Paper and Answer key Download PDF. 10thh Maths Unit 2 Numbers and Sequence TN 10th All Subject July Month Assignment 2 Answers, 10th Tamil, English, Maths, Science, Social Science July Month Assignment Answer key Tamil Medium, English Medium Download PDF. TN SCERT,  ASSIGNMENT 1 QUESTION & ANSWER KEY TN SCERT


  • 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,nЄN, 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 F= Fn-1 + Fn-2,ifF1=1, F2=3, then Fis
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, ar…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 (13+23+33 …+153) – (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)
10TH MATHS ASSIGNMENT



14. Given that
10th Maths July Month Assignment Answer key
10th Maths July Month Assignment Answer key


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 …
July Month Assignment Answer key


17. If 1+2+3+…+n= 666 then find n.

10th Maths July Month Assignment Answer key



Part – C

III. Write in details.

18. Find the sum to n terms of the series 5 + 55 + 555 …

Find the sum to n terms of the series 5 + 55 + 555 …
19. Find the HCF of 396, 504, 636

10th Maths July Month Assignment Answer key
Find the HCF of 396, 504, 636


20. In an AP, the sum of four consecutive terms is 28 and the sum of their squares is 276. Find that four numbers.

In an AP, the sum of four consecutive terms is 28 and the sum of their squares is 276. Find that four numbers.



21. Rekha has 15 square colour papers of size 10cm, 11 cm, 12cm …. 24cm. How much area can be decorated with these colour papers?

10cm, 11 cm, 12cm …. 24cm.

Post a Comment

0 Comments