# How do you identify the 15th term of the geometric sequence 52, 39, 29.25 …?

${t}_{15} \approx 0.926$
${t}_{n} = {t}_{1} \cdot {r}^{n - 1}$
${t}_{1} = 52 , r = \frac{39}{52} = \frac{3}{4}$
${t}_{15} = 52 \cdot {\left(\frac{3}{4}\right)}^{14}$
${t}_{15} \approx 0.926$