Question

In an AP with a = 3 and d = 12, find the term that will be 120 more than it's ${21}^{(}st)$ term.

Answer 1

a = 3
d = 12
$a_n=a_{21}+120$
= a + 20s + 120
= 3 + 20$\times$12 + 120
= 123 + 240
= 363
363 = 3 +(n-1)$\times$12
$n-1=\frac{360}{12}$
n = 30 + 1 = 31
${31}^{st}$ term of the given A.P. is 120 more than it's ${21}^{st}$ term.