Algebra II / Mr. Hansen

Name: _________________________________

S1.

t₂₀ = 34(1.7)¹⁹ = 812,846.2813
S₂₀ = 34(1-rⁿ)/(1-r) = 34(1-1.7²⁰)/(1-1.7) » 34(-40641.31407)/(-0.7) » 1,974,006.683

S2.

t₃₀ = 78+29(3.2) = 170.8
S₃₀ = 30/2 (t₁ + t₃₀) = 15(78+170.8) = 3,732

S3.

You may be able to see that d = 11 without any work. Consequently, t₂ = 55.
S₁₀₀ requires finding t₁₀₀ first. We can use d = 11 to get t₁₀₀ = 44 + 99(11) = 1133. Therefore, S₁₀₀ = 58,850 by the formula in the blue box on p.583. (Note: You could also use the blue box at the top of p.584 to skip the step of calculating t₁₀₀, but then you have a more complicated formula to remember and apply.)

S4.

We can compute r by dividing any term by the one just before it. We are given two terms, t₁ and t₂, so divide t₂ by t₁ to get 80/100 = 0.8 = r. Therefore, we have these answers:
t₃ = 80(0.8) = 64
t₄ = 64(0.8) = 51.2
t₅ = 51.2(0.8) = 40.96
Finally, does it make sense to talk about the partial sums converging to a limit? Well, yes, since |r| < 1. (Note: You should always check this before using the infinite sum formula.) Now we can go ahead and use the formula on p.591: S = t₁/(1-r) = 100/(1-0.8) = 100/0.2 = 500.

S5.

This problem is similar to S4, except that r = 0.95 instead of 0.8. It's a geometric sequence (note: we don't say "series" this time since there's no discussion of partial or infinite sums). By the formula at the bottom of p.567, t₁₀ = 800(0.95)⁹ » 504.1995278.

S6.

This is an arithmetic sequence with t₁ = 900, d = -7, so t₁₀ = 900 + 9(-7) = 837. To find the first negative term, solve the inequality t_n < 0 for n. This gives 900 + (n-1)(-7) < 0 Þ 907-7n < 0 Þ -7n < -907 Þ n > 129.571428, so you must go out to the 130th term to get a negative term.

S7.

Multiplying by r a total of 9 times (i.e., r⁹) takes us from 12 to 120, so 12r⁹ = 120 Þ r⁹ = 10 Þ r = 10^1/9 » 1.29155. To reach 1200 we must multiply t₁₀ (namely, 120) by 10, but 10 equals r⁹ as we saw a moment ago. Thus we must go 9 terms beyond t₁₀, i.e., to the 19th term. If you're not comfortable reasoning like this, you could also use "pure algebra" to solve the equation t_n = 12rⁿ-1 = 1200 for n. This yields rⁿ-1 = 100 Þ (10^1/9)ⁿ-1 = 100 Þ 1/9 (n-1) = 2 Þ n-1 = 18 Þ n = 19.

S8.

Since t₁₀ = 27, and t₁₀ = t₁ + 9d, we have t₁ + 9d = 27 Þ 17 + 9d = 27 Þ 9d = 10 Þ d = 10/9. Therefore, t₂₀ = t₁ + 19d = 17 + 19(10/9) = 343/9 » 38.111.

S9.

5/6, 6/7; neither arithmetic nor geometric (no common difference, no common ratio)

S10.

1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 = 10! = 3,628,800

1...

Don't let the word "increase" fool you into thinking this problem is asking about the total size of your biceps. We have a geometric sequence of "increase" amounts, not total amounts. Let t₁ = increase on day 1, let t₂ = increase on day 2, and so on. In effect, the problem tells you that t₁ = 3 and r = 0.95.

a.

t₁₀ = 3(0.95)⁹ = 1.890748229 » 1.89 mm

b.

S₁₀ = 3(1-r¹⁰)/(1-r) = 3(0.4012630608)/0.05 » 24.1 mm

c.

Check: |r| = 0.95 < 1, so we can proceed. Infinite sum is S = 3/(1-r) = 3/0.05 = 60 mm.

3.a.

If 4% of the color is lost each time, then 96% of the color remains each time (compared to the time before). This is a sequence where each term is 96% as large as the one before it. In other words, we have a geometric sequence with r = 0.96. Let t₁ = fraction of color left after 1 washing, let t₂ = fraction left after 2 washings, and so on. The problem tells you indirectly that t₁ = 96%, which we can write as 0.96.

b.

t₁₀ = 0.96(0.96)⁹ » 0.664832636, or about 66.5%

c.

Solve the inequality t_n £ 0.25 for n. Can you do this? By the formula at bottom of p.567, t_n = t₁(rⁿ-1) = 0.96(0.96)ⁿ-1 = 0.96ⁿ, and this final expression must be £ 0.25. Take logs of both sides of our new inequality (namely 0.96ⁿ £ 0.25) to get n log 0.96 £ log 0.25, or n ³ (log 0.25)/(log 0.96). By the way, why does the direction of the inequality flip when we divide both sides by log 0.96? This final expression is approximately 33.96, and since n must be the next integer that is ³ 33.96, we take n = 34 washings.

d.

An arithmetic sequence (with fixed d) would predict that the fraction of color remaining keeps decreasing at a constant linear rate. But this makes no sense; the fraction can't be negative, as this model would eventually predict.

4...

Let t₁ = distance traveled between 1st and 2nd bounce, let t₂ = distance traveled between 2nd and 3rd bounce, etc. In other words, we will let t_n = distance traveled between the nth and (n+1)st bounce.

a.

t₁ = 36 m (reason: the height is 90% of 20 m, or 18 m, and the ball must travel up to that height and then back down)

b.

Each bounce is 90%, 0.9, as high as the one before. Thus we have a geometric sequence with r = 0.9.

c.

t₆ = 36(0.9)⁵ = 21.25764 » 21.26 m

d.

Since the question says between, we go up to (but not including) the 7th bounce. Answer: S₆ = 36(1-rⁿ)/(1-r) = 36(1-0.9⁶)/(1-0.9) = 36(0.468559)/(0.1) = 168.68124 » 168.68 m.

6...

Let t₁ = amount owed after one full day of this arrangement, t₂ = amount owed after 2 days, etc. The problem tells us that t₁ = 0.10+5%(0.10) = 0.105 (i.e., 10 and a half cents). We know (see Example 3 on p.600) that this is a geometric sequence with r = 1.05. Reason: Each day, the balance is the old balance multiplied by 105% (that includes the 5% interest), and 105% is just another way of saying 1.05.

a.

t₃₆₅ = 0.105(1.05)³⁶⁴ » $5,421,184.16

b.

t₃₆₆ = 0.105(1.05)³⁶⁵ » $5,692,243.37, which is $271,059.21 more for a single day!

14...

Let t₁ = amount the parents pay on day 1, let t₂ = amount the parents pay on day 2, etc. We can see that this is a geometric sequence with t₁ = 0.01 and r = 2. After 30 days, those poor, gullible, misguided parents will have paid S₃₀ = 0.01(1-rⁿ)/(1-r) = 0.01(1-2³⁰)/(1-2) = 0.01(-1,073,741,823)/(-1) = $10,737,418.23. Warning: If you get your parents to go for this idea, I cannot assume any responsibility!

First, note that 4% annual interest can be written as 0.04. Divide by 12 to get the interest rate in each month, namely 0.04/12. Add this to 100% (i.e., 1) to get the total factor that the account balance is multiplied by each month. (There's 100% of the money, plus 4%/12 for the monthly interest.) This factor, namely, 1.0033333, is r. In this problem, the rest of the calculations should be performed using months, not years.

a.

After 1 month, we have $20,000(1.0033333), or $20,066.67, and this is our t₁. From this we can get the amount after 10 years (i.e., 120 months), which is t₁₂₀ = 20,066.67(1.0033333)¹¹⁹ » $29,816.65, and the amount after 50 years (i.e., 600 months), which is t₆₀₀ = 20,066.67(1.0033333)⁵⁹⁹ » $147,290.43.

b.

After 1 month, we have $100(1.0033333), or $100.33, and this is the first term of a new sequence, or actually we should say series since we'll be adding up all these $100 payments with their interest. Let u₁ = balance at end of 1st month, u₂ = balance at end of 2nd month, etc. We use u for the terms, instead of t, to avoid confusion with the sequence in part (a). After 10 years (120 months), our balance using the method of $100 monthly payments is S₁₂₀ = 100.33333(1-rⁿ)/(1-r) = 100.33333(1-1.0033333¹²⁰)/(1-1.0033333) » 100.33333(-0.4908267)/(-0.0033333) » $14,774 to the nearest dollar, but since another $100 payment (without interest) arrives on the first of the month, we have $14,874 total after 10 years. After 50 years (600 months), our balance is S₆₀₀ = 100.33333(1-1.0033333⁶⁰⁰)/(1-1.0033333) » 100.33333(-6.364375)/(-0.0033333) » $191,572 to the nearest dollar, but again we must add $100 for the newly arriving payment to get $191,672 total after 50 years.

c.

You could solve an inequality for n, but that is rather tedious. You could also use the sequence mode of your TI-83, but that is even worse, since you have 600 data points to plot. Instead, I recommend using your TI-83 and pretending that the expressions for t_n in part (a) and S_n in part (b) are really continuous functions of x instead of discrete functions of n.

Begin by "STO"ing 1+.04/12 into R so that R is 1.0033333. (This makes sense, because you're keeping 100%, or simply 1, plus a monthly percentage that is 1/12 of the annual percentage.)

Since t₁ is 20000*R, we can write t_n as t₁rⁿ-1 = 20000R(R^(N-1)) = 20000R^N, or using our "function of x" trick, Y₁ = 20000R^X. Actually press your "Y=" key and punch this function in, please. Similarly, punch in Y₂ = 100R(1-R^X)/(1-R)+100. (Why does Y₂ have a "+100" at the end?) Now graph with a window of [0, 600] × [0, 200000]. You'll see that the second curve overtakes the first curve about midway through the 50-year (600-month) span. Use 2nd CALC to find the actual intersection point, namely X=__________. Therefore, the second curve will exceed the first curve after 328 months (27.25 years). If you think you will live that long, you should choose the $100-per-month payout instead of the lump sum.

Note: Strictly speaking, you should not rely on smooth curves, since the domain for both functions is {natural numbers}, not  . However, the smooth curves are much easier to work with than sequences in this case. In a problem involving a smaller domain (say, the integers from 1 to 25), the TI-83 "Seq" mode would be more reasonable. However, you have to define partial sums in a special way on your calculator, and we probably won't have time to discuss this in class. (For advanced students: Use a recursive sequence definition to handle the partial sums.)

d.

Change R by "STO"ing 1+.07/12 into R. Then do a ZOOM ZoomFit to redraw the curves. Can you see that this time, the first curve starts higher and increases faster than the second curve? In other words, if the assumed interest rate is 7% instead of 4%, the $100-per-month payout plan can never catch up with the lump sum plan. For advanced students: One can show (via tricky algebra) that an annual interest rate of 5.6745% or more prevents the $100-per-month payout plan from catching up with the lump sum plan in the first 50 years. By using some even trickier algebra and taking the limit as x (or n) grows without bound, one can show that an interest rate of 6.0302% or more prevents the $100-per-month payout plan from ever catching up with the lump sum plan.

e.

If you consider the effects of annual income tax paid from the proceeds of the account, you have a lower net interest rate. For example, if you live in Virginia and have a federal tax rate of 28% and a state tax rate of 5.75%, you have to pay more than a third of your interest earnings as tax, and if the nominal rate is 4%, your actual after-tax rate is only 2.65%. Since lower interest rates make it harder for the lump sum plan to maintain its lead, the effect of taxes could very well change the conclusions we reached in parts (c) and (d). If you really want to be sophisticated, you could compare the two payout plans under various tax deferral or tax avoidance strategies, such as a 401(k) plan or a Roth IRA.