OMPT sequences and series practice: lesson and 10 problems

Answer: 83

The common difference is $d=11-7=4$, and the first term is $a_1=7$. The nth term is $a_n=a_1+(n-1)d$, so $a_{20}=7+19\cdot 4=7+76=83$. Note the 19: walking from term 1 to term 20 takes nineteen steps, not twenty.

Answer: Step 2

Step 1 quotes the formula correctly and Step 3 solves the equation it was handed. The substitution in Step 2 is what breaks: with $n=5$ the multiplier is $n-1=4$, so the equation is $23=a_1+4\cdot 3$ and $a_1=11$. Check by walking forward: $11,14,17,20,23$. Five terms, four steps.

Answer: 820

Here $a_1=3$ and $d=4$. The last term is $a_{20}=3+19\cdot 4=79$, and the sum of an arithmetic series is the average of the first and last term times the number of terms: $S_{20}=20\cdot \frac{3+79}{2}=20\cdot 41=820$. The same thing written as one formula is $S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)=10(6+76)=820$.

Answer: B — $39366$

The ratio is $r=6÷2=3$ and $a_1=2$, so $a_n=a_1{r}^{n-1}$ gives $a_{10}=2\cdot 3^9=2\cdot 19683=39366$. The exponent is $n-1=9$ for exactly the reason the arithmetic formula uses $n-1$: term one has been multiplied by $r$ nine times to become term ten.

Answer: 1476

Each year the output is multiplied by $0.8$, so the four yearly figures are a geometric sequence with $a_1=500$ and $r=0.8$: 500, 400, 320, 256. The finite geometric sum is $S_n=a_1\frac{1-r^n}{1-r}$, so $S_4=500\cdot \frac{1-0.8^4}{0.2}=500\cdot \frac{1-0.4096}{0.2}=500\cdot 2.952=1476$. Adding the four numbers by hand gives 1476 as well, which is the check worth doing when the ratio is friendly.

Answer: 13.5

The ratio is $r=3÷9=\frac{1}{3}$, and $|r|<1$, so the infinite sum exists and equals $S=\frac{a_1}{1-r}=\frac{9}{1-\frac{1}{3}}=\frac{9}{\frac{2}{3}}=\frac{27}{2}=13.5$. Sanity check: the first four terms already add to $13\frac{1}{3}$, so a total just above 13.5 would be impossible and just under it is exactly right.

Answer: C — $12-8+\frac{16}{3}-\frac{32}{9}+\cdots$

An infinite geometric series converges exactly when $|r|<1$. The first has $r=\frac{3}{2}$, too big. The second has $r=-1$, so $|r|=1$ and the partial sums bounce between 5 and 0 without settling. The fourth is not geometric at all; it is arithmetic with $d=1$ and it grows without bound. The third has $r=-\frac{2}{3}$, so $|r|=\frac{2}{3}<1$ and the sum is $\frac{12}{1+\frac{2}{3}}=\frac{36}{5}=7.2$.

Answer: 354

From term 4 to term 9 is five steps, so $5d=42-17=25$ and $d=5$. Stepping back three from $a_4$ gives $a_1=17-3\cdot 5=2$. Then $S_{12}=\frac{12}{2}\left(2\cdot 2+11\cdot 5\right)=6(4+55)=6\cdot 59=354$. Check with the other form: $a_{12}=2+11\cdot 5=57$, and $12\cdot \frac{2+57}{2}=354$.

Answer: 210

Written out, the terms are $1,4,7,\dots ,34$: an arithmetic series with $a_1=1$, $d=3$ and $n=12$ terms. So the sum is $12\cdot \frac{1+34}{2}=6\cdot 35=210$. Splitting the sigma works too: $3\sum k-\sum 2=3\cdot 78-24=234-24=210$, using ${\sum }_{k=1}^{12}k=\frac{12\cdot 13}{2}=78$.

Answer: Step 2

Step 1 reads the series correctly and Step 3 does its arithmetic correctly. Step 2 is where the reasoning fails: $S=\frac{a_1}{1-r}$ is only valid when $|r|<1$, and here $r=3$. The terms are growing, the partial sums run off to infinity, and the correct answer is that the series has no sum. The negative result is the giveaway: a series of positive terms can never add to $-1$.