How do you find A1 in a geometric sequence if A3=-8 and A6=1?

Nov 8, 2015

${A}_{1} = - 32$

Explanation:

${A}_{n} = {A}_{1} {r}^{n - 1}$

${A}_{3} = - 8 = {A}_{1} {r}^{3 - 1}$

$\implies - 8 = A 1 {r}^{2}$

${A}_{6} = 1 = {A}_{1} {r}^{6 - 1}$

$\implies 1 = {A}_{1} {r}^{5}$

$\implies 1 = {A}_{1} {r}^{2} {r}^{3}$

But $- 8 = A 1 {r}^{2}$

$\implies 1 = - 8 {r}^{3}$

$\implies - \frac{1}{8} = {r}^{3}$

$\implies r = - \frac{1}{2}$

${A}_{3} = {A}_{1} {r}^{3 - 1}$

$\implies - 8 = {A}_{1} \cdot {\left(- \frac{1}{2}\right)}^{2}$

$\implies - 8 = {A}_{1} \cdot \frac{1}{4}$

$\implies - 32 = {A}_{1}$