# In each of the ar

Let a = first term of arithmetic sequence

a2 , a3 , a4… are second, third, fourth terms and so on

d = common difference

i) Here, a = 24

a2 = 42

d = 42-24 = 18

a3 = 42 + 18 = 60

a4 = 60 + 18 = 78

ii) Here, a2 = 24

a3 = 42

d = 42-24 = 18

a = a2 – d = 24-18 = 6

a4 = 42 + 18 = 60

iii) Here, a3 = 24

a4 = 42

d = a4 – a3 = 42-24 = 18

a2 = 24 – 18 = 6

a = 6-18 = -12

iv) Here, a = 24

a3 = 42

a3 = a + 2d

Solving, d = 9

a2 = 24 + 9 = 33

a4 = 42 + 9 = 51

v) Here, a2 = 24

a4 = 42

a4 = a + 3d …(eq)1

and,a2 = a + d …(eq)2

Eliminating a;

d = 9

a = 24-9 = 15

a3 = 24 + 9 = 33

vi) Here, a = 24

a4 = 42

a4 = a + 3d

Solving, d = 6

a2 = 24 + 6 = 30

a3 = 30 + 6 = 36

Hence, all the answers are found.

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