Question

Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21^st term?

Answer 1

In the given problem, let us first find the 21^st term of the given A.P.

A.P. is 3, 15, 27, 39 …

Here

First term (a) = 3

Common difference of the A.P. (d) = 15 - 3 = 12

Now, as we know,

`a_n = a + (n - 1)d`

So for 21^st term (n = 21)

`a_21 = 3 + (21 - 1)(12)`

= 3 + 20(12)

= 3 + 240

= 243

Let us take the term which is 120 more than the 21^st term as `a_n` So, 

`a_n = 120 + a_21`

`a_21 = 3 + (21 - 1)(12)`

`= 3+ 20(12)`

= 3 + 240

= 243

Let us take the term which is 120 more than the 21^st term as `a_n` so

`a_n = 120 + a_21`

= 120 + 243

= 363

Also `a_n = a + (n - 1)d`

363 = 3 + (n - 1)12

363 = 3 + 12n - 12

363 + 9 = 12n

Further simplifying we get

372 = 12n

`n = 372/12`

n = 31

Therefore the 31^st term of the given A.P is 120 more than the 21^st term.