**(1) 1, 8, 15, 22, . . .**

**Solution:**

Given 1, 8, 15, 22, . . .

First term a = 1

Second term t_{1} = 8

Third term t_{2} = 15

Fourth term t_{3} = 22

We know that d = t_{n + 1} – t_{n}

Thus, t_{2} – t_{1} = 15 – 8 = 7

t_{3} – t_{2} = 22 – 15 = 7

Thus, d = 7

**(2) 3, 6, 9, 12, . . .**

**Solution:**

Given 3,6,9,12, . . .

First term a = 3

Second term t_{1} = 6

Third term t_{2} = 9

Fourth term t_{3} = 12

We know that d = t_{n + 1} – t_{n}

Thus, t_{2} – t_{1} = 9 – 6 = 3

t_{3} – t_{2} = 12 – 9 = 3

Thus, d = 3

**(3) – 3, – 8, – 13, – 18, . . .**

**Solution:**

Given – 3, – 8, – 13, – 18, . . .

First term a = – 3

Second term t_{1} = – 8

Third term t_{2} = – 13

Fourth term t_{3} = – 18

We know that d = t_{n + 1} – t_{n}

Thus, t_{2} – t_{1} = – 13 – (– 8) = – 13 + 8 = – 5

t_{3} – t_{2} = – 18 – (– 13) = – 18 + 13 = – 5

Thus, d = – 5

**(4) 70, 60, 50, 40, . . .**

**Solution:**

Given 70, 60, 50, 40, . . .

First term a = 70

Second term t_{1} = 60

Third term t_{2} = 50

Fourth term t_{3} = 40

We know that d = t_{n + 1} – t_{n}

Thus, t_{2} – t_{1} = 50 – 60 = – 10

t_{3} – t_{2} = 40 – 50 = – 10

Thus, d = – 10