# 10th CBSE math solution for exercise 5.1 part 2

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## 10th CBSE math solution for exercise 5.1 part 2

(2) Write first four terms of the A.P when the first term a and common difference d are given as follows.

(i) a = 1 0  d = 10

Solution:

First term (a) = 10

Second term (a₂) = a + d

= 10 + 10

= 20

Third term (a₃) = a + 2d

= 10 + 2(10)

= 10 + 20

= 30

Fourth term (a₄) = a + 3d

= 10 + 3(10)

= 10 + 30

= 40

Therefore the first four terms are 10,20,30 and 40

(ii) a = -2  d = 0

Solution:

First term (a) = -2

Second term (a₂) = a + d

= -2 + 0

= -2

Third term (a₃) = a + 2d

= -2 + 2(0)

= -2 + 0

= -2

Fourth term (a₄) = a + 3d

= -2 + 3(0)

= -2 + 0

= -2

Therefore the first four terms are -2,-2,-2 and -2

(iii) a = 4  d = -3

Solution:

First term (a) = 4

Second term (a₂) = a + d

= 4 + (-3)

= 1

Third term (a₃) = a + 2d

= 4 + 2(-3)

= 4 - 6

= -2

Fourth term (a₄) = a + 3d

= 4 + 3(-3)

= 4 - 9

= -5

Therefore the first four terms are 4,1,-2 and -5

(iv) a = -1  d = 1/2

Solution:

First term (a) = -1

Second term (a₂) = a + d

= -1 + (1/2)

= -1/2

Third term (a₃) = a + 2d

= -1 + 2(1/2)

= -1 + 1

= 0

Fourth term (a₄) = a + 3d

= -1 + 3(1/2)

= -1 + (3/2)

= 1/2

Therefore the first four terms are -1,-1/2,0 and 1/2

(v) a = - 1.25   d = -0.25

Solution:

First term (a) = -1.25

Second term (a₂) = a + d

= -1.25 + (-0.25)

= -1.25 - 0.25

= -1.50

Third term (a₃) = a + 2d

= -1.25 + 2(-0.25)

= -1.25 - 0.50

= -1.75

Fourth term (a₄) = a + 3d

= -1.25 + 3(-0.25)

= -1.25 - 0.75

= 2

Therefore the first four terms are -1.25,-1.50,-1.75 and 2