Answer :

Consider the AP 3, 15, 27, 39, ……
We know that the nth term of an AP is given by:

an = a + (n – 1)d

The 21st term will be:

a21 = 3 + (21 – 1)12
⇒ a21 = 3 + (20)12
⇒ a21 = 3 + 240
⇒ a21 = 243
According to the question,

an = 120 + a21

⇒a + (n – 1)d = 120 + 243

⇒3 + (n – 1)12 = 363
⇒ 3 + 12n - 12 = 363
⇒ 12n - 9 = 363
⇒ 12n = 372

⇒ n = 31

So, 31st term of the AP is 120 more than its 21st term i.e. 363.

OR

Sn = 3n2 – 4n

At n = 1, S1 = -1

At n = 2, S2 = 4

At n = 3, S3 = 15

We know that an AP is also represented by sum of nth terms.

AP = S1, (S2 – S1), (S3 – S2), ……
AP = -1,(4+1),(15-4), .......  

AP = -1, 5, 11, ……

an = a + (n – 1)d

an = -1 + (n – 1)6
= -1 + 6n - 6
= 6n – 7

So, nth term of an AP is given by 6n – 7.

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