Question

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1. An arithmetic sequence that has t(28) = 6 and t(22) = 42. Write a rule to describe the sequence.

2. Kim invested $5000 in a Mutual fund at 7.6% compounded quarterly.
a. Write an equation to represent this situation
b. When will her investment be worth more than $10,000.

Answer 1

`\textsf{1.} \quad a_n=174-6n`

`\textsf{2.\;a)} \quad A=5000\left(1.019\right)^(4t)`

b) 9.25 years (to the nearest quarter)

Explanation:

Question 1

`\boxed{\begin{minipage}{8 cm}\underline{General form of an arithmetic sequence}\\\\$a_n=a+(n-1)d$\\\\where:\\\phantom{ww}$\bullet$ $a_n$ is the nth term. \\ \phantom{ww}$\bullet$ $a$ is the first term.\\\phantom{ww}$\bullet$ $d$ is the common difference between terms.\\\phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}`

Given terms of an arithmetic sequence:

• 
  `a_(28)=6`

• 
  `a_(22)=42`

Substitute the given values into the arithmetic sequence formula to create two equations:

`\implies a_(28)=a+27d=6`

`\implies a_(22)=a+21d=42`

Subtract the second equation from the first equation to eliminate a and solve for d:

`\implies 6d=-36`

`\implies d=-6`

Substitute the found value of d into one of the equations and solve for a:

`\implies a+27(-6)=6`

`\implies a-162=6`

`\implies a=168`

Substitute the found values of a and d into the formula to create an equation for the nth term of the sequence:

`\implies a_n=168+(n-1)(-6)`

`\implies a_n=168-6n+6`

`\implies a_n=174-6n`

Question 2

`\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\$ A=P\left(1+(r)/(n)\right)^(nt)$\\\\where:\\\\ \phantom{ww}$\bullet$ $A =$ final amount \\ \phantom{ww}$\bullet$ $P =$ principal amount \\ \phantom{ww}$\bullet$ $r =$ interest rate (in decimal form) \\ \phantom{ww}$\bullet$ $n =$ number of times interest is applied per year \\ \phantom{ww}$\bullet$ $t =$ time (in years) \\ \end{minipage}}`

Given values:

• P = $5000
• r = 7.6% = 0.076
• n = 4 (quarterly)

Substitute the given values into the compound interest formula to create an equation with respect to time, t:

`\implies A=5000\left(1+(0.076)/(4)\right)^(4t)`

`\implies A=5000\left(1.019\right)^(4t)`

To find when the value of the investment will be worth more than $10,000, substitute A = 10000 into the equation:

`\implies 10000 < 5000\left(1.019\right)^(4t)`

`\implies 2 < \left(1.019\right)^(4t)`

`\implies \ln 2 < \ln\left(1.019\right)^(4t)`

`\implies \ln 2 < 4t\ln\left(1.019\right)`

`\implies (\ln 2)/(4 \ln\left(1.019\right)) < t`

`\implies t > 9.20672924...`

Therefore, the investment will be worth more than $10,000 from 9.25 years (to the nearest quarter).