Question

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

Answer 1

In the given problem, let us first find the 21st 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),
a₂₁ = 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.

a_n = 120 + a₂₁
So, a_n = 120 + 243
a_n = 363

Also, a_n = a +(n - 1)d
363 = 3 + (n -1)12
363 = 3 + 12n - 12
363 = -9 = 12n

363+ 9 = 12n

Further simplifying, we get
372 = 12n 

`"n" = (370)/(12)`

n = 31

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