Answer :

Formula Used.

d_{n} = a_{n + 1} – a_{n}

a_{n} = a + (n–1)d

In the above sequence,

a = 12;

d_{1} = a_{2}–a_{1} = 17–12 = 5

d_{2} = a_{3}–a_{2} = 22–17 = 5

d_{3} = a_{4}–a_{3} = 27–22 = 5

The difference in sequence is same and comes to be (5).

For any term of A.P to be 0

a_{n} = a + (n–1)d = 0

a_{n} = a + (n–1)d = 12 + (n–1)(5)

= 12 + 5n–5

= 7 + 5n

7 + 5n = 0

5n = –7

n =

∴ The number of terms cannot be negative

Hence, the no term of A.P can be 0

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