# In a geometric sequence, a_3=-9 and a_6=243, find the value of a_10?

Mar 17, 2018

The value of ${a}_{10} = 19683$

#### Explanation:

a_3=-9 , a_6=243 ; a_10= ?

$n$ th term in G.P series is a_n=a_1*r^(n-1) ; a_1 and r are

first term and common ratio of G.P. series.

a_3=a_1*r^(3-1) or a_1*r^2= -9 ; (1) similarly ,

a_6=a_1*r^(6-1) or a_1*r^5= 243 ; (2) Dividing

equation (2) by equation (1) we get, ${r}^{5} / {r}^{2} = - \frac{243}{9}$ or

${r}^{3} = - 27 \mathmr{and} r = - 3$ Putting $r = - 3$ in equation (1) we

get , ${a}_{1} \cdot {\left(- 3\right)}^{2} = - 9 \mathmr{and} {a}_{1} = - \frac{9}{9} \mathmr{and} {a}_{1} = - 1$

$\therefore {a}_{10} = {a}_{1} \cdot {r}^{10 - 1} = - 1 \cdot {\left(- 3\right)}^{9} = 19683$

The value of ${a}_{10} = 19683$ [Ans]

Mar 17, 2018

$19 , 683$

#### Explanation:

The general term for a GP is ${a}_{n} = {a}_{1} {r}^{n - 1}$

where ${a}_{1}$ is the first term and $r$ is the common ratio.

You are given the values of two terms in a GP.

Divide the two terms: The formula and the values

${a}_{6} / {a}_{3} = \frac{\cancel{a} {r}^{5}}{\cancel{a} {r}^{2}} = \frac{243}{-} 9$

This gives: ${r}^{3} = - 27 \text{ } \leftarrow$ find the cube root.

$\textcolor{w h i t e}{\times \times \times \times} r = - 3$

Now find ${a}_{1}$ from ${a}_{3}$

${a}_{1} {r}^{2} = {a}_{1} {\left(- 3\right)}^{2} = - 9$

${a}_{1} = \frac{- 9}{9} = - 1$

Now you can find the value of ${a}_{10}$

${a}_{10} = - 1 {\left(- 3\right)}^{9}$

$= 19 , 683$