Answer :

a = 3, d = 15 – 3 = 12

Let last term be a_{n}

a_{n} = a + (n – 1) d

= 3 + (n – 1) 12

= 12n – 9 (i)

a_{21} = a + 20d

= 3 + 20(12) = 243

Now, the term is 120 more than the 21^{st} term

a_{n} = 120 + a_{21}

= 120 + 243

= 363

Putting this value in (i), we get

363 = 12n – 9

12n = 363 + 9

n = 31

Hence, 31^{st} term of given A.P. is 120 more than its 21^{st} term

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