Question

Find how many terms of a geometric progression 1+3+9 are required to make a total of more than 1 million.

Answer 1

The partial sum of a geometric sequence is

$\displaystyle \sum_{i=0}^N a^i = \dfrac{a^{N+1}-1}{a-1}$

In your case a=3, so if we sum N terms of the sequence we have

$\displaystyle \sum_{i=0}^N 3^i = \dfrac{3^{N+1}-1}{2}$

We want this to me more than 1 million, so we have

$\dfrac{3^{N+1}-1}{2}>1000000 \iff 3^{N+1}-1>2000000 \iff 3^{N+1} > 1999999$

Considering the log (base 3) of both sides, we have

$N+1>\log_3(1999999)\iff N>\log_3(1999999)-1 approx 12.2$

So, starting from N=13, the sum of the first N terms will be more than 1 million