Arithmetic 3

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  1. Here, first term a1 = 10 and common difference, d = 10

Hence,

2nd term a2 = a1 + d

= 10 + 10

= 20

3rd term a3 = a1 + 2d

= 10 + 2 x 10

= 30

4th term a4 = a1 + 3d

= 10 + 30

= 40

Therefore,

first four terms of the AP are:

10, 20, 30, 40, ……

  1. Here,

First term a = 1 and Common difference = 0

Therefore, first four terms of the given AP are:

a1 = - 2, a2 = - 2, a3 = - 2 and a4 = - 2

  1. Here, first term a1 = 4 and common difference d = - 3

We know that an = a + (n – 1)d, where n = number of terms

Thus, second term a2 = a + (2 – 1)d

a2 = 4 + (2-1)×(-3)

= 4 - 3 = 1

3rd term a3 = a + (3 – 1)d

 = 4 + (3-1) × (-3)

= 4 - 6

= -2

4th term a4 = a + (4-1)d

= 4 + (4 - 1) ×( -3)

= 4 - 9 = -5

Therefore,

First four terms of given AP are:

4, 1, - 2, - 5

  1. We have,

1st term = - 1 and d = ½

Hence,

2nd term a2 = a1 + d

                  = -1 + ½

                  = - ½

3rdterm a3 = a1 + 2d

                 = -1 + 2 * ½

                 = 0

4th term a4 = a1 + 3d

                 = -1 + 3 * ½

                 = ½

Therefore,

The four terms of A.P. are -1, - ½, 0, ½

 

  1. We have

1st term = - 1.25 and d = - 0.25

2nd term a2 = a + d

                  = -1.25 - 0.25

                  = -1.5

3rd term a3 = a + 2d

                   = -1.25 + 2 × (-0.25)

                   = -1.25 - 0.5

                   = -1.75

4th term a4 = a + 3d

                  = 1.25 + 3 × (-0.25)

                  = -2.25

Therefore, first four terms of the A.P. are: 1.25, -1.5, -1.75 and – 2.25

 

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