How do you find the 7th term of the geometric sequence with the given terms a4 = -4, a6 = -100?

Aug 2, 2018

$\pm 500$.

Explanation:

In the Usual Notation, ${a}_{n} = {a}_{1} {r}^{n - 1} , n \in \mathbb{N}$.

Given that, ${a}_{6} = - 100 \mathmr{and} {a}_{4} = - 4$.

$\Rightarrow {a}_{1} {r}^{5} = - 100 \mathmr{and} {a}_{1} {r}^{3} = - 4$.

$\therefore \frac{{a}_{1} {r}^{5}}{{a}_{1} {r}^{3}} = \frac{- 100}{- 4} = 25$.

$\therefore {r}^{2} = 25.$

$\therefore r = \pm 5$.

r=-5, &, a_4=-4 rArr a_1=-4/r^3=(-4)/(-5)^3=4/5^3.

Then, ${a}_{6} = {a}_{1} {r}^{5} = \left(\frac{4}{5} ^ 3\right) \cdot {\left(- 5\right)}^{5} = - 100$.

Hence, in this case, ${a}_{7} = {a}_{1} {r}^{6} = \left(\frac{4}{5} ^ 3\right) \cdot {\left(- 5\right)}^{6} = 500$.

In case, r=+5, then, a_1=-4/5^3, &, a_6=-4/5^3*5^5=-100

Then, ${a}_{7} = {a}_{6} \cdot r = \left(- 100\right) \left(+ 5\right) = - 500$.

Aug 2, 2018

${T}_{7} = \pm 500$

Explanation:

We know that any term in a geometric sequence can be described as ${T}_{n} = a {r}^{n - 1}$
where
$a$ is the first term
$n$ is the nth term
$r$ is the ratio between 2 adjacent terms

If we know that ${T}_{4} = - 4$ and ${T}_{6} = - 100$ and ${T}_{n} = a {r}^{n - 1}$, we can solve to find $a$ and $r$

${T}_{4} = a {r}^{4 - 1} = - 4$
${T}_{4} = a {r}^{3} = - 4$ ---- (1)

${T}_{6} = a {r}^{6 - 1} = - 100$
${T}_{6} = a {r}^{5} = - 100$ ---- (2)

$\frac{\left(2\right)}{\left(1\right)}$

$\frac{a {r}^{5}}{a {r}^{3}} = - \frac{100}{-} 4$

${r}^{2} = 25$

$r = \pm 5$

If we know that $r = 5$, then subbing $r = 5$ back into (1)
$a {\left(5\right)}^{3} = - 4$
$125 a = - 4$
$a = - \frac{4}{125}$

To test if it is correct, sub $a = - \frac{4}{125}$ into (2)

$L H S$
$= - \frac{4}{125} \times {5}^{5}$
$= - \frac{4}{125} \times 3125$
$= - 100$
$= R H S$

If we know that $r = - 5$, then subbing $r = - 5$ back into (1)
$a {\left(- 5\right)}^{3} = - 4$
$- 125 a = - 4$
$a = \frac{4}{125}$

To test if it is correct, sub $a = \frac{4}{125}$ into (2)

$L H S$
$= \frac{4}{125} \times {\left(- 5\right)}^{5}$
$= \frac{4}{125} \times - 3125$
$= - 100$
$= R H S$

Therefore, we know that $r = \pm 5$ and $a = \pm \frac{4}{125}$
To find ${T}_{7}$,
${T}_{7} = \pm \frac{4}{125} \times {5}^{7 - 1} = \pm \frac{4}{125} \times {5}^{6} = \pm 500$

Aug 2, 2018

$\text{The "7^(th)"term of geometric sequence is:}$

${a}_{7} = 500 \mathmr{and} - 500$

Explanation:

We know that,

color(green)(n^(th) "term of the Geometric sequence is :"

color(green)(a_n=a_1(r)^(n-1)

where ,color(green)(a_1="first term" and r="common ratio."

We have,

a_4=-4color(white)(;;;;.............;;;;)and a_6=-100

:.a_4=a_1(r)^(4-1)=-4 color(white)(;;;)and a_6=a_1(r)^(6- 1)=-100

$\therefore {a}_{1} {r}^{3} = - 4. \ldots \to \left(1\right) \mathmr{and} {a}_{1} {r}^{5} = - 100. . . \to \left(2\right)$

$e q n . \left(2\right) \implies {a}_{1} {r}^{3} \cdot {r}^{2} = - 100$

$\implies \left(- 4\right) {r}^{2} = - 100. . . \to u s e , e q n \left(1\right)$

$\implies {r}^{2} = \frac{- 100}{- 4} = 25$

color(red)(=>r+-5

Now ,

(i)"for " color(red)( r=5

${7}^{t h} t e r m = {a}_{7} = {a}_{1} {\left(r\right)}^{7 - 1}$

$\implies {a}_{7} = {a}_{1} {r}^{6} = {a}_{1} {r}^{5} \cdot {r}^{1}$

$\implies {a}_{7} = \left(- 100\right) \left(+ 5\right) \to \left[\because e q n . \left(2\right) \mathmr{and} \textcolor{red}{r = 5}\right]$

=>color(blue)(a_7=-500

(i)"for " color(red)( r=-5

${7}^{t h} t e r m = {a}_{7} = {a}_{1} {\left(r\right)}^{7 - 1}$

$\implies {a}_{7} = {a}_{1} {r}^{6} = {a}_{1} {r}^{5} \cdot {r}^{1}$

$\implies {a}_{7} = \left(- 100\right) \left(- 5\right) \to \left[\because e q n . \left(2\right) \mathmr{and} \textcolor{red}{r = - 5}\right]$

=>color(blue)(a_7=500