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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.

Respuesta :

zayaanedu

Answer:

Question 1:

An arithmetic sequence is a sequence where the difference between any two consecutive terms is constant. If t(28) = 6 and t(22) = 42, we can determine the common difference by subtracting the two terms: 42 - 6 = 36. To find the first term of the sequence, we can use the formula for the nth term of an arithmetic sequence: t(n) = a + (n-1)d, where a is the first term, d is a common difference, and n is the position of the term in the sequence. Plugging in the given values, we have:

t(28) = a + (28-1)36

6 = a + 27(36)

6 = a + 972

a = -966

So the rule to describe the sequence is t(n) = -966 + (n-1)36

Question 2:

a. The equation to represent the situation is: A = P(1+r/n)^(nt) where A is the amount of the investment, P is the principal amount, r is the interest rate, n is the number of times the interest is compounded in a year and t is the number of years the investment is held.

so the equation will be A = 5000(1+0.076/4)^(4*t)

b. To find when the investment is worth more than $10,000, we need to find when A > 10000. We can set A = 10000 and solve for t.

A = 5000(1+0.076/4)^(4t)

10000 = 5000(1+0.076/4)^(4t)

2 = (1+0.076/4)^(4*t)

t = (log2)/(log(1+0.076/4))/4

so t is the number of years it will take for the investment to be worth more than $10,000

Step-by-step explanation:

semsee45

Answer:

[tex]\textsf{1.} \quad a_n=174-6n[/tex]

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

   b)  9.25 years (to the nearest quarter)

Step-by-step explanation:

Question 1

[tex]\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}}[/tex]

Given terms of an arithmetic sequence:

  • [tex]a_{28}=6[/tex]
  • [tex]a_{22}=42[/tex]

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

[tex]\implies a_{28}=a+27d=6[/tex]

[tex]\implies a_{22}=a+21d=42[/tex]

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

[tex]\implies 6d=-36[/tex]

[tex]\implies d=-6[/tex]

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

[tex]\implies a+27(-6)=6[/tex]

[tex]\implies a-162=6[/tex]

[tex]\implies a=168[/tex]

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

[tex]\implies a_n=168+(n-1)(-6)[/tex]

[tex]\implies a_n=168-6n+6[/tex]

[tex]\implies a_n=174-6n[/tex]

Question 2

[tex]\boxed{\begin{minipage}{8.5 cm}\underline{Compound Interest Formula}\\\\$ A=P\left(1+\frac{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}}[/tex]

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:

[tex]\implies A=5000\left(1+\dfrac{0.076}{4}\right)^{4t}[/tex]

[tex]\implies A=5000\left(1.019\right)^{4t}[/tex]

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

[tex]\implies 10000 < 5000\left(1.019\right)^{4t}[/tex]

[tex]\implies 2 < \left(1.019\right)^{4t}[/tex]

[tex]\implies \ln 2 < \ln\left(1.019\right)^{4t}[/tex]

[tex]\implies \ln 2 < 4t\ln\left(1.019\right)[/tex]

[tex]\implies \dfrac{\ln 2}{4 \ln\left(1.019\right)} < t[/tex]

[tex]\implies t > 9.20672924...[/tex]

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

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