Chapter 5 Arithmetic Progressions - Ncert Solutions for Class 10 Mathematics CBSE

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Chapter 5 - Arithmetic Progressions Excercise Ex. 5.1

Solution 1
Solution 2
Solution 3
Solution 4

Chapter 5 - Arithmetic Progressions Excercise Ex. 5.2

Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20

Chapter 5 - Arithmetic Progressions Excercise Ex. 5.3

Solution 1

(i)     2, 7, 12 ,…, to 10 terms

For this AP,

a = 2

d = a2a1 = 7 – 2 = 5 

n = 10

We know that,

(ii)     –37, –33, –29 ,…, to 12 terms

For this AP,

a = –37

d = a2a1 = (–33) – (–37)

 = – 33 + 37 = 4 

n = 12

We know that,


(iii)   0.6, 1.7, 2.8 ,…, to 100 terms

For this AP,

a = 0.6

d = a2a1 = 1.7 – 0.6 = 1.1  

n = 100

We know that,



 


Solution 2
Solution 3



Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10

(i)

(ii)

Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20

Chapter 5 - Arithmetic Progressions Excercise Ex. 5.4

Solution 1
Solution 2
Solution 3
Solution 4
Solution 5