Sequences, Series, and Interest Practice

These questions follow Sequences and Their Rules, Series, and Money Over Time — one road from patterns to dollars. A scientific calculator earns its keep from question 7 onward; money answers to the nearest cent.

Sequences

1. Classify each as arithmetic, geometric, or neither, with the evidence: (a) $5,9,13,17,\dots$ (b) $4,12,36,108,\dots$ (c) $1,4,9,16,\dots$
2. For the arithmetic sequence $7,10,13,\dots$: write the general term, then find $t_{20}$.
3. A geometric sequence has $t_1=3$ and $r=2$. Write the general term and find $t_{10}$.
4. (a) Write a recursion formula for $2,6,18,54,\dots$ (b) Write the first four terms of $t_1=5$, $t_n=t_{n-1}-4$.

Answer 1

(a) Arithmetic — constant difference $d=4$. (b) Geometric — constant ratio $r=3$. (c) Neither — the differences $3,5,7$ grow, and the ratios shrink. (They are the perfect squares: a discrete quadratic, not a member of either family.)

Answer 2

$t_n=7+(n-1)(3)=3n+4$. Then $t_{20}=3(20)+4=64$ — the general term teleports; no need to climb through nineteen steps.

Answer 3

$t_n=3\cdot 2^{n-1}$. Then $t_{10}=3\cdot 2^9=3(512)=1536$. Note the exponent is $n-1$: the first term has been multiplied zero times.

Answer 4

(a) $t_1=2$, $t_n=3t_{n-1}$ — starting value plus growth rule; a recursion without its first term is a rule with nowhere to stand. (b) $5,1,-3,-7$ — arithmetic, with $d=-4$.

Series

5. Find the sum of the first 40 terms of $3+7+11+\cdots$
6. Find the sum of the first 10 terms of $2+6+18+\cdots$

Answer 5

Arithmetic, $a=3$, $d=4$: $S_{40}=\frac{40}{2}\left[2(3)+39(4)\right]=20(162)=3240$. Gauss’s pairing in formula form — forty additions traded for one multiplication.

Answer 6

Geometric, $a=2$, $r=3$: $S_{10}=\frac{2\left(3^{10}-1\right)}{3-1}=3^{10}-1=59,048$. Notice the last term is only $2\cdot 3^9=39,366$ — in geometric series, the final term carries most of the total.

Money

7. $1000 is invested for 10 years at 6% per year. Find the amount if interest compounds (a) annually and (b) monthly, and state how much the extra compounding earned.
8. An investment earns 8% per year, compounded annually. Use systematic guess-and-check to find how many years it takes to double.
9. You deposit $100 at the end of every month for 5 years into an account earning 6% per year, compounded monthly. Find the future value, and how much of it is interest.
10. A $2000 credit-card balance sits unpaid for two years at 20% per year, compounded monthly. What is the debt then, and what did the waiting cost?

Answer 7

(a) $A=1000(1.06{)}^{10}\approx \text{textdollar}1790.85$. (b) $i=0.005$, $n=120$: $A=1000(1.005{)}^{120}\approx \text{textdollar}1819.40$. Monthly compounding earned an extra $28.55 — same rate on paper, more meetings with the multiplier.

Answer 8

Solve $(1.08{)}^n=2$ by hunting: $(1.08{)}^5\approx 1.47$, $(1.08{)}^9\approx 1.999$, $(1.08{)}^{10}\approx 2.16$. Doubling takes almost exactly 9 years. Every guess was cheap; the strategy — bracket the target, then close in — is the reusable part.

Answer 9

An ordinary annuity: $R=100$, $i=0.005$, $n=60$, so $FV=\frac{100\left[(1.005{)}^{60}-1\right]}{0.005}\approx \text{textdollar}6977.00$. You deposited $60\times \text{textdollar}100=\text{textdollar}6000$; the other $977.00 is interest — the geometric series doing the saving alongside you.

Answer 10

$i=\frac{0.20}{12}$, $n=24$: $A=2000{\left(1+\frac{0.20}{12}\right)}^{24}\approx \text{textdollar}2973.86$. Waiting cost about $974 — nearly half the original debt again, in two years. The same exponential that grew question 9’s savings works just as tirelessly for the lender.