How many terms of the series r 3^(1/2) + 3 + 3*(3^(1/2)) + 9

We have a following sequence: \(\sqrt{3}\), \(3\), \(3\sqrt{3}\), \(9\), ...

Notice that this sequence is a geometric progression with first term equal to \(\sqrt{3}\) and common ratio also equal to \(\sqrt{3}\).

The question asks: the sum of how many terms of this sequence adds up to \(120+121\sqrt{3}\).

The sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}-1)}{r-1}\), (where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\)).

So, \(\frac{\sqrt{3}(\sqrt{3}^n-1)}{\sqrt{3}-1}=120+121\sqrt{3}\) --> \(\sqrt{3}(\sqrt{3}^n-1)=120\sqrt{3}+121*3-120-121\sqrt{3}\) --> \(\sqrt{3}*\sqrt{3}^n-\sqrt{3}=243-\sqrt{3}\) --> \(\sqrt{3}*\sqrt{3}^n=243\) --> \(3^{\frac{1+n}{2}}=3^5\) --> \(\frac{1+n}{2}=5\) --> \(n=9\).

Answer: E.

Shortcut solution:
Alternately you can spot that every second term (which also form a geometric progression), 3, 9, 27, ... should add up to 120: \(\frac{3(3^k-1)}{3-1}=120\) --> \(3^k-1=80\) --> \(3^k=81\) --> \(k=4\), so the number of terms must be either \(2k=8\) (if there are equal number of irrational and rational terms) or \(2k+1=9\) (if # of irrational terms, with \(\sqrt{3}\), is one more than # rational terms), since only 9 is present among options then it must be a correct answers.

Hope it's clear.