Solve the problem.
-Find the nth term of the following geometric sequence: $1 + \sqrt { 3 } , 3 + \sqrt { 3 } , 3 + 3 \sqrt { 3 } , \ldots$
A) $( 1 + \sqrt { 3 } ) ( \sqrt { 3 } ) ^ { n }$
B) $( 3 + \sqrt { 3 } ) ( \sqrt { 3 } ) ^ { n - 1 }$
C) $( 1 + \sqrt { 3 } ) ( \sqrt { 3 } ) ^ { \mathrm { n } - 1 }$
D) $( 1 + \sqrt { 3 } ) \cdot 3 ^ { n - 1 }$

Question

Quizplus · Accepted Answer

The common ratio of the sequence can be found by dividing the second term by the first term: $\frac{3 + \sqrt{3}}{1 + \sqrt{3}} = \sqrt{3}$. The first term is $1 + \sqrt{3}$, so the nth term is given by the first term multiplied by the common ratio raised to the power of $n-1$, which matches choice C: $(1 + \sqrt{3})(\sqrt{3})^{n-1}$.

Solve the Problem 1+3,3+3,3+33,…1 + \sqrt { 3 } , 3 + \sqrt { 3 } , 3 + 3 \sqrt { 3 } , \ldots1+3​,3+3​,3+33​,…

Solve the Problem $1 + \sqrt { 3 } , 3 + \sqrt { 3 } , 3 + 3 \sqrt { 3 } , \ldots$