Q:

Geometric Progression related question.

In a sequence, first term a1 = 100 and nth term, an = 100 + (an-1)/5 If for some integer k, a50 lies between k and (k + 1), then k =

A:

Let's check the first 3 terms:
$a_1 = 100$
$a_2 = 100 + 100/5 = 120$
$a_3 = 100 + 100/5 + 100/25 = 124$
I don't see how this is an arithmetic progression. It's more like a geometric series:
$a_n = \sum_{i=0}^{n-1} 100*(1/5)^i$
This allows a simple geometric series formula. Can you solve it from here?

