Answer :

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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