# Question 36bf7

Mar 12, 2016

${n}^{\text{th}}$ $\text{term} = {t}_{n} = 2 {\left(3\right)}^{n - 1}$
number of terms$= 7$

#### Explanation:

Recall that a geometric sequence can be written in the following form:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{n} = a {r}^{n - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
${t}_{n} =$term number
$a =$first term
$r =$common ratio
$n =$number of terms

Question 1 i)
$1$. To determine the ${n}^{\text{th}}$ term, we must form an equation for the geometric sequence. Start by finding the value of $r$, the common ratio. To do this, divide any term by the term before it. In this case, we will divide the second term by the first term.

$r = {t}_{2} / {t}_{1}$

$r = \frac{6}{2}$

$r = 3$

$2$. Substitute the values of $a$ and $r$ into the geometric sequence formula.

${t}_{n} = a {r}^{n - 1}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {t}_{n} = 2 {\left(3\right)}^{n - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\therefore$, the ${n}^{\text{th}}$ term is ${t}_{n} = 2 {\left(3\right)}^{n - 1}$.

Question 1 ii)
$1$. To determine the number of terms in the sequence, take the formula for the general term derived in part i) and substitute 1458, the last term, as ${t}_{n}$.

${t}_{n} = 2 {\left(3\right)}^{n - 1}$

$1458 = 2 {\left(3\right)}^{n - 1}$

$2$. Solve for $n$.

$729 = {3}^{n - 1}$

${3}^{6} = {3}^{n - 1}$

$6 = n - 1$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} n = 7 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\therefore$, there are $7$ terms in the sequence.

Mar 12, 2016

$a = 4$
$r = 2$

#### Explanation:

Question 2
$1$. Using the general formula for a geometric sequence, ${t}_{n} = a {r}^{n - 1}$, substitute the second term, $8$, and the fifth term, $64$, along with their other appropriate known values into two separate equations.

${t}_{n} = a {r}^{n - 1} \textcolor{w h i t e}{X X X X X X X X x} {t}_{n} = a {r}^{n - 1}$

$8 = a {r}^{2 - 1} \textcolor{w h i t e}{X X X X X X X X X} 64 = a {r}^{5 - 1}$

$8 = a r \textcolor{w h i t e}{X X X X X X X X X X X} 64 = a {r}^{4}$

$2$. Using elimination, divide $64 = a {r}^{4}$ by $8 = a r$ to find the value of $r$.

$\frac{64 = a {r}^{4}}{8 = a r}$

$\frac{64 = \textcolor{red}{\cancel{\textcolor{b l a c k}{a}}} {r}^{4}}{8 = \textcolor{red}{\cancel{\textcolor{b l a c k}{a}}} r}$

$8 = {r}^{3}$

$r = \sqrt[3]{8}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} r = 2 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$3$. Knowing that $r = 2$, substitute the value into $8 = a r$ to find the value of $a$.

$8 = a r$

$8 = a \left(2\right)$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} a = 4 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\therefore$, the first term is $4$ and the common ratio is $2$.

Mar 12, 2016

$a = 6$
$r = \frac{1}{3}$

#### Explanation:

Question 3
$1$. Since the first term is represented by $\textcolor{\mathmr{and} a n \ge}{a}$, the second term by $\textcolor{b r o w n}{a r}$, the third term by $\textcolor{p u r p \le}{a {r}^{2}}$, and the fourth term by $\textcolor{b l u e}{a {r}^{3}}$, use these terms to create two algebraic expressions.

Equation $1$: $\textcolor{\mathmr{and} a n \ge}{a} + \textcolor{p u r p \le}{a {r}^{2}} = \frac{20}{3}$

Equation $2$: $\textcolor{b r o w n}{a r} + \textcolor{b l u e}{a {r}^{3}} = \frac{20}{9}$

$2$. For each equation, solve for $a$.

Equation $1$:$\textcolor{w h i t e}{X X X X X X X X X}$Equation $2$:

$\textcolor{\mathmr{and} a n \ge}{a} + \textcolor{p u r p \le}{a {r}^{2}} = \frac{20}{3} \textcolor{w h i t e}{X X X X X X X} \textcolor{b r o w n}{a r} + \textcolor{b l u e}{a {r}^{3}} = \frac{20}{9}$

$a \left(1 + {r}^{2}\right) = \frac{20}{3} \textcolor{w h i t e}{X X X X X X} a \left(r + {r}^{3}\right) = \frac{20}{9}$

$a = \frac{20}{3 \left(1 + {r}^{2}\right)} \textcolor{w h i t e}{X X X X X X X} a = \frac{20}{9 \left(r + {r}^{3}\right)}$

$3$. Since you rearranged the two equations in terms of $a$, then the equations must equal each other. Thus, set the two equations equal to each other and solve for $r$.

$\frac{20}{\textcolor{t u r q u o i s e}{3} \left(1 + {r}^{2}\right)} = \frac{20}{\textcolor{t e a l}{9} \left(r + {r}^{3}\right)}$

$\left(\textcolor{t u r q u o i s e}{3} \cdot \textcolor{t e a l}{9}\right) \left[\frac{20}{\textcolor{t u r q u o i s e}{3} \left(1 + {r}^{2}\right)}\right] = \left(\textcolor{t u r q u o i s e}{3} \cdot \textcolor{t e a l}{9}\right) \left[\frac{20}{\textcolor{t e a l}{9} \left(r + {r}^{3}\right)}\right]$

$\left(\textcolor{red}{\cancel{\textcolor{t u r q u o i s e}{3}}} \cdot \textcolor{t e a l}{9}\right) \left[\frac{20}{\textcolor{red}{\cancel{\textcolor{t u r q u o i s e}{3}}} \left(1 + {r}^{2}\right)}\right] = \left(\textcolor{t u r q u o i s e}{3} \cdot \textcolor{red}{\cancel{\textcolor{t e a l}{9}}}\right) \left[\frac{20}{\textcolor{red}{\cancel{\textcolor{t e a l}{9}}} \left(r + {r}^{3}\right)}\right]$

$\frac{9 \left(20\right)}{1 + {r}^{2}} = \frac{3 \left(20\right)}{r + {r}^{3}}$

$\frac{180}{\textcolor{m a \ge n t a}{1 + {r}^{2}}} = \frac{60}{r \textcolor{m a \ge n t a}{\left(1 + {r}^{2}\right)}}$

$\textcolor{m a \ge n t a}{\left(1 + {r}^{2}\right)} \left(\frac{180}{\textcolor{m a \ge n t a}{1 + {r}^{2}}}\right) = \textcolor{m a \ge n t a}{\left(1 + {r}^{2}\right)} \left[\frac{60}{r \textcolor{m a \ge n t a}{\left(1 + {r}^{2}\right)}}\right]$

$\textcolor{red}{\cancel{\textcolor{m a \ge n t a}{\left(1 + {r}^{2}\right)}}} \left(\frac{180}{\textcolor{red}{\cancel{\textcolor{m a \ge n t a}{1 + {r}^{2}}}}}\right) = \textcolor{red}{\cancel{\textcolor{m a \ge n t a}{\left(1 + {r}^{2}\right)}}} \left[\frac{60}{r \textcolor{red}{\cancel{\textcolor{m a \ge n t a}{\left(1 + {r}^{2}\right)}}}}\right]$

$180 = \frac{60}{r}$

$180 r = 60$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} r = \frac{1}{3} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$4$. Knowing that $r = \frac{1}{3}$, substitute this value into either equation $1$ or $2$ to solve for $a$. In this case, we will use equation $1$.

$a = \frac{20}{3 \left(1 + {r}^{2}\right)}$

a=20/(3(1+(1/3)^2)#

$a = \frac{20}{3 \left(1 + \frac{1}{9}\right)}$

$a = \frac{20}{3 \left(\frac{9 + 1}{9}\right)}$

$a = \frac{20}{{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}^{1} \left(\frac{10}{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}} ^ 3\right)}$

$a = \frac{20}{\frac{10}{3}}$

$a = 20 \cdot \frac{3}{10}$

$a = {\textcolor{red}{\cancel{\textcolor{b l a c k}{20}}}}^{2} \cdot \frac{3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{10}}}} ^ 1$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} a = 6 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\therefore$, $a$ is $6$ and $r$ is $\frac{1}{3}$.