Question

which term of the AP 3, 15, 27, 39 so on will be 120 more than its 21st term

Answer 1

your answer is :-

Given AP: 3 , 15,27,39

such that a = 3 & d = 12

Let , nth term be its 120 more than its 21st term

•°• Tn = 120 + T21

=> a + (n-1)d = 120 + a + 20d

=> 3 + 12n - 12 = 120 + 3 + 20×12

=> 12n - 12 = 360

=> n = 372/12

=> n = 31

hence , 31 term is 120 more than its 21st term

Answer 2

Heya...!!!

_Here Is Your Answer_

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In the given AP
first term , a = 3
and ,
common difference , d = a2 - a1
d = 15 - 3
d = 12
the 21st term of the AP is
= a + 20 d
= 3 + 20 × 12
= 3 + 240
= 243
the term which is more than 120 of the 21 st term is = 243 + 120 = 363
Now ,
an = a + (n - 1 ) d
363 = 3 + (n - 1 ) 12
363 - 3 = (n - 1 ) 12
360 / 12 = n - 1
30 + 1 = n
n = 31
so , the 31st term of the given AP is more than 120 of its 21st term

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