# In a geometric sequence t_3 = 45 t_6=1215, how do you find the first term?

Nov 10, 2015

${t}_{1} = 5$

#### Explanation:

For any geometric sequence, the common ratio between successive terms has value $r = \frac{{t}_{n} + 1}{t} _ n$.

From this we can conclude that ${t}_{n} = {r}^{n - 1} \cdot {t}_{1}$

Hence if we work from term 3 to term 6 we may write that

${r}^{3} {t}_{3} = {t}_{6}$

$\therefore 45 {r}^{3} = 1215$

$\therefore r = \sqrt[3]{\frac{1215}{45}} = 3$.

This implies that ${t}_{1} = {t}_{3} / {r}^{2} = \frac{45}{9} = 5$