# How do you write a rule for the nth term of the geometric term and then find a_6 given r=-1/2, a_3=8?

Jun 14, 2017

${a}_{n} = {a}_{1} \cdot {r}^{n - 1} \mathmr{and} {a}_{6} = - 1$

#### Explanation:

Rule of finding n th term of G.P is ${a}_{n} = {a}_{1} \cdot {r}^{n - 1}$ .where ${a}_{1} , r , n$
are 1st term, common ratio and required term respectively.

a_3=8 , r=-1/2, a_6= ?

${a}_{3} = {a}_{1} \cdot {\left(- \frac{1}{2}\right)}^{3 - 1} \mathmr{and} 8 = {a}_{1} \cdot {\left(- \frac{1}{2}\right)}^{2} \mathmr{and} 8 = {a}_{1} \cdot \frac{1}{4}$ or

a_1=32 ; :. a_6= a_1* (-1/2)^(6-1) or a_6 = 32* (-1/2)^5 or

${a}_{6} = 32 \cdot \left(- \frac{1}{32}\right) = - 1$

${a}_{6} = - 1$ [Ans]

Jun 14, 2017

${a}_{n} = 32 {\left(- \frac{1}{2}\right)}^{n - 1} , \mathmr{and} , {a}_{6} = - 1.$

#### Explanation:

In the Usual Notation for Geometric Sequence,

we are given, $r = - \frac{1}{2} , \mathmr{and} , {a}_{3} = 8. \ldots \left(1\right) .$

Knowing that, ${a}_{n} = {a}_{1} \cdot {r}^{n - 1} , \text{ we have, using } \left(1\right) ,$

$8 = {a}_{3} = {a}_{1} \cdot {\left(- \frac{1}{2}\right)}^{3 - 1} , \mathmr{and} , {a}_{1} = 32.$

$\therefore {a}_{n} = 32 {\left(- \frac{1}{2}\right)}^{n - 1} .$

$\therefore {a}_{6} = 32 {\left(- \frac{1}{2}\right)}^{5} = - 1.$