Answer :

Formula Used.

d_{n} = a_{n + 1} – a_{n}

a_{n} = a + (n–1)d

In the above sequence,

a = 17;

d_{1} = a_{2}–a_{1} = 22–17 = 5

d_{2} = a_{3}–a_{2} = 27–22 = 5

d_{3} = a_{4}–a_{3} = 32–27 = 5

⇒ As in A.P the difference between the 2 terms is always constant

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

∴ The above sequence is A.P

The n^{th} term of A.P is a_{n} = a + (n–1)d

a_{n} = a + (n–1)d = 17 + (n–1)5

= 17 + 5n–5

= 12 + 5n

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