Question

quantos números existem de 15 até 329?

Answer 1

an=a1+(n-1)r
329=15+(n-1)
n-1=329-15
n-1=314
n=315

Answer 2

an = a1 + ( n -1) . r

329 = 15 + ( n -1) . 1

329 = 15 + 1n - 1

329 = 14 + 1n

315 = 1n

n = 315