Answer :

Consider the AP 3, 15, 27, 39, ……

We know that the n^{th} term of an AP is given by:

a_{n} = a + (n – 1)d

^{st}term will be:

a_{21} = 3 + (21 – 1)12

⇒ a_{21} = 3 + (20)12

⇒ a_{21} = 3 + 240

⇒ a_{21} = 243

According to the question,

a_{n} = 120 + a_{21}

⇒a + (n – 1)d = 120 + 243

⇒3 + (n – 1)12 = 363

⇒ 3 + 12n - 12 = 363

⇒ 12n - 9 = 363

⇒ 12n = 372

⇒ n = 31

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

**OR**

S_{n} = 3n^{2} – 4n

At n = 1, S_{1} = -1

At n = 2, S_{2} = 4

At n = 3, S_{3} = 15

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

AP = S_{1}, (S_{2} – S_{1}), (S_{3} – S_{2}), ……

AP = -1,(4+1),(15-4), .......

AP = -1, 5, 11, ……

a_{n} = a + (n – 1)d

a_{n} = -1 + (n – 1)6

= -1 + 6n - 6

= 6n – 7

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

Rate this question :