Which term of the sequence \( 1 ; 3 ; 9 ; \ldots \) is equal to 729 ?

To find which term of the sequence \( 1 ; 3 ; 9 ; \ldots \) is equal to 729, we first recognize the sequence as a geometric progression where the first term \( a = 1 \) and the common ratio \( r = 3 \). The \( n \)-th term of a geometric sequence can be described by the formula: \[ a_n = a \cdot r^{(n-1)} \] Substituting the values we have: \[ a_n = 1 \cdot 3^{(n-1)} = 3^{(n-1)} \] We want to find \( n \) such that \( a_n = 729 \): \[ 3^{(n-1)} = 729 \] Next, we recognize that 729 can be expressed as a power of 3: \[ 729 = 3^6 \] Now, we can set the exponents equal to each other: \[ n - 1 = 6 \] Solving for \( n \) gives: \[ n = 6 + 1 = 7 \] Thus, the term of the sequence that is equal to 729 is the **7th** term.