Arithmetic Sequences & Series - Algebra 3

A theater has 18 seats in the first row and each successive row has three more seats than the previous row. If the last row has 72 seats, what is the total number of seats in the theater?
Answer: 855 The row count is $n=\frac{72-18}{3}+1=19$ . The total is $S_{19}=\frac{19}{2}(18+72)=19\cdot45=855$ . The values 900, 945, and 990 come from counting 20, 21, or 22 rows, typical errors when handling the inclusive endpoints.
What is the sum of all integers from $-13$ through $47$ , inclusive?
Answer: $1037$ When you see "sum of all integers from $a$ through $b$ , inclusive," you're finding an arithmetic series. The key formula is: $\text{count} = b - a + 1, \quad \text{sum} = \text{count} \times \frac{a + b}{2}.$ Here $a=-13$ , $b=47$ , so the count is $47 - (-13) + 1 = 61,$ and the average of the first and last terms is $\frac{-13 + 47}{2} = \frac{34}{2} = 17.$ Thus, $61 \times 17 = 1037.$ That is the correct sum. Notice the pair structure: the 61 terms have middle value $17$ , so your sum is $30$ pairs of sum $34$ plus the middle $17$ , which also gives $30 \times 34 + 17 = 1037$ . The choice $799$ comes from using $47$ as the number of terms: $47 \times 17 = 799$ . But $47$ is just the endpoint, not the count — the count includes the negative numbers and zero, so it must be $61$ . The choice $1020$ is $60 \times 17$ . That happens if you compute the count as $47 - (-13) = 60$ , forgetting to add $1$ for "inclusive." You need that extra term, which is the middle $17$ . The choice $1054$ is $62 \times 17$ . That trap comes from adding $1$ twice — treating both endpoints as extra. But the difference $60$ already measures gaps; for inclusive endpoints, add exactly one, giving $61$ . Study takeaway: for any consecutive integer sum, always write the count first: $\text{last} - \text{first} + 1$ . If your count times the average is around $1000$ , you can sanity-check against $61 \times 17$ .
An arithmetic sequence has first term $4$ and ninth term $52$ . What is the sum of the terms from the 5th term through the 15th term, inclusive?
Answer: $638$ The common difference is $\frac{52-4}{9-1}=6$ , so $a_n=4+(n-1)6=6n-2$ . Thus $a_5=28$ and $a_{15}=88$ , with $15-5+1=11$ terms in the sum. Therefore $\frac{11}{2}(28+88)=11\cdot58=638$ . $580$ uses $10/2$ , not $11/2$ , as the multiplier for $(28+88)$ ; $610$ is $S_{15}-S_5$ , which excludes the 5th term; $690$ is $S_{15}$ , forgetting to subtract the first four terms.
For the arithmetic sequence $a_n=7-2n$ , which formula gives the sum of the first $n$ terms?
Answer: $n(6-n)$ The first term is $a_1=5$ , and the $n$ th term is $7-2n$ . Therefore $S_n=\frac n2(5+7-2n)=\frac n2(12-2n)=n(6-n)$ . The formula $n(7-n)$ results from using $7$ as the first term; $n(5-n)$ results from using $5-2n$ as the $n$ th term; $n(6+n)$ results from a sign error when simplifying the average.
In the arithmetic sequence $a_n = 3n + 2$ , what is the sum of the terms from $a_6$ through $a_{10}$ , inclusive?
Answer: $130$ Whenever you need the sum of consecutive terms in an arithmetic sequence, your first move is to identify the endpoints and the number of terms. For an inclusive range, the count is last index minus first index plus one. Here, $a_6 = 3(6)+2 = 20$ and $a_{10}=3(10)+2=32$ . From $a_6$ through $a_{10}$ inclusive there are $10-6+1=5$ terms, and the average of the first and last terms is $\frac{20+32}{2}=26$ . So the sum is $5\cdot 26=130$ . The distractor $156$ comes from an off-by-one error: if you mistakenly count 6 terms, $6\cdot 26=156$ . Remember that $a_6, a_7, a_8, a_9, a_{10}$ is only five terms. The choice $185$ is the sum of the first ten terms, $a_1$ through $a_{10}$ , since $\frac{10(5+32)}{2}=185$ ; that includes $a_1$ through $a_5$ , which are not wanted. The choice $55$ is the sum of the first five terms, $a_1$ through $a_5$ , because $\frac{5(5+17)}{2}=55$ ; it starts at the wrong endpoint. For inclusive sums, always calculate the term count as $n = \text{last} - \text{first} + 1$ . If the range is small, listing the terms quickly ( $20+23+26+29+32$ ) can also help you avoid endpoint errors.
In an arithmetic sequence, $a_1=5$ and $S_8=124$ . What is the common difference $d$ ?
Answer: $3$ An arithmetic sequence is linear: each term is the previous term plus $d$ . When you know the first term, the number of terms, and the sum, the key formula is $S_n=\frac{n}{2}(2a_1+(n-1)d).$ Plug in $n=8$ , $a_1=5$ , and $S_8=124$ : $124=\frac{8}{2}(2(5)+7d)=4(10+7d)=40+28d.$ So $84=28d$ , giving $d=3$ . The common difference is therefore $3$ . The choice $31$ is a trap: since $S_8=4(a_1+a_8)$ , you get $a_1+a_8=31$ , but that is the sum of the first and eighth terms, not the common difference. The choice $\frac{21}{8}$ comes from using $a_8=a_1+8d$ instead of $a_1+7d$ , which gives $124=40+32d$ . The choice $\frac{13}{4}$ often comes from taking that $31$ as if it were $a_8$ , then computing $(31-5)/8=26/8=13/4$ . That compounds two mistakes: $31$ is not $a_8$ , and there are only $7$ gaps from $a_1$ to $a_8$ . Remember: with $n$ terms, the last term uses $n-1$ differences. Also, $\frac{2S_n}{n}$ gives $a_1+a_n$ , not $a_n$ alone. Check your indexing before dividing.
The second term of an arithmetic sequence is $9$ , and the fifth term is $21$ . What is the sum of the first 15 terms?
Answer: $495$ An arithmetic sequence is built by adding the same difference $d$ each time. You are told the second term is $9$ and the fifth term is $21$ , so between those two terms there are $5-2=3$ steps. That means $21-9=12$ is equal to $3d$ , so $d=4$ . Work backward to the first term: since the second term is $9$ , the first term is $9-4=5$ . Now use the sum formula for the first 15 terms: $S_{15}=\frac{15}{2}\left[2(5)+(15-1)(4)\right] =\frac{15}{2}(10+56) =\frac{15}{2}(66)=495.$ So $495$ is the correct sum. The choice $555$ comes from treating the second term, $9$ , as if it were the first term and then summing 15 terms with that starting value: $\frac{15}{2}(18+14\cdot4)=555$ . The choice $525$ incorrectly combines the second term with the correct 15th term, using an average of $\frac{9+61}{2}$ , which mixes up which term is the true starting point. The choice $465$ would require the first term to be $3$ , but then the second term would be $7$ , not $9$ ; this is an off-by-one indexing error. The key is always to identify $a_1$ and $d$ before summing. Count the intervals between known terms carefully, then use $S_n=\frac n2(a_1+a_n)$ .
An employee's first-year salary is $50{,}000$ . If salaries increase by a fixed amount each year and total earnings over the first 10 years are $635{,}000$ , what is the annual increase?
Answer: $3000$ The sum formula gives $S_{10}=\frac{10}{2}[2(50{,}000)+9d]=500{,}000+45d$ . Set this equal to $635{,}000$ , so $45d=135{,}000$ and $d=3000$ . The values $2000$ , $2500$ , and $3500$ would produce totals $590{,}000$ , $612{,}500$ , and $657{,}500$ , respectively, so they correspond to common errors in handling the number raises or the sum formula.
The first four terms of an arithmetic sequence are $3,7,11,15$ . How many terms, beginning with the first, must be added so that the sum is $903$ ?
Answer: 21 Here $d=4$ , so $S_n=\frac n2[2(3)+(n-1)4]=n(2n+1)$ . Setting $n(2n+1)=903$ gives $2n^2+n-903=0$ . The discriminant is $85^2$ , so $n=\frac{-1+85}{4}=21$ . The values 18, 19, and 20 correspond to sums 666, 741, and 820, not 903.
An arithmetic series has first term $17$ , last term $-13$ , and sum $30$ . What is the common difference?
Answer: $-\frac{30}{14}$ Since $S_n=\frac n2(17-13)=2n$ and $S_n=30$ , $n=15$ . There are $n-1=14$ intervals, so $d=\frac{-13-17}{14}=\frac{-30}{14}$ . The other denominators come from using 13, 15, or 16 intervals or terms incorrectly.
The sum of the first 10 terms of an arithmetic sequence is $230$ and the sum of the first 20 terms is $860$ . What is the 15th term?
Answer: $61$ From $S_{10}=5(2a_1+9d)=230$ , $2a_1+9d=46$ . From $S_{20}=10(2a_1+19d)=860$ , $2a_1+19d=86$ . Subtracting gives $10d=40$ , so $d=4$ and $a_1=5$ . Thus $a_{15}=5+14(4)=61$ . The value $63$ is $\frac{S_{20}-S_{10}}{10}$ , the average of terms 11 through 20, not $a_{15}$ ; $65$ is $a_{16}$ , and $69$ is $a_{17}$ .
An arithmetic sequence has first term $-3$ . If the sum of its first 15 terms is $480$ , what is the common difference?
Answer: 5 Using $S_{15}=\frac{15}{2}[2(-3)+14d]=480$ , we get $-45+105d=480$ , so $d=5$ . The values $4.375$ , $4.667$ , and $5.385$ come from counting 16, 15, or 13 intervals between the first and 15th terms instead of 14 intervals.
What is the sum of all positive even integers less than $200$ that are not divisible by $4$ ?
Answer: $5000$ All even integers less than 200 form $2,4,\ldots,198$ : 99 terms, sum $\frac{99}{2}(2+198)=9900$ . The multiples of 4 less than 200 form $4,8,\ldots,196$ : 49 terms, sum $\frac{49}{2}(4+196)=4900$ . Subtracting the excluded multiples gives $9900-4900=5000$ . The value $4900$ is the excluded sum, $9900$ is all evens less than 200, and $10100$ is all evens up to and including 200.
The 9th term of an arithmetic sequence is $-14$ , and the 17th term is $26$ . Which formula gives the nth term of this sequence?
Answer: $a_n=5n-59$ Whenever a problem gives you two terms of an arithmetic sequence, your job is to find the common difference $d$ and then the starting term $a_1$ . Here the 17th and 9th terms are separated by $17-9=8$ positions, and the value increases by $26-(-14)=40$ . So $d=\frac{40}{8}=5$ . Now use the 9th term to step back to $a_1$ : $a_9=a_1+8(5)$ , so $-14=a_1+40$ , which gives $a_1=-54$ . Thus the nth term is $a_n=a_1+(n-1)d=-54+5(n-1)=5n-59$ . Plugging in $n=9$ gives $-14$ , and $n=17$ gives $26$ , so this formula matches the data. Each wrong choice has slope $5$ but the wrong intercept. The formula $5n-54$ comes from writing $a_n=-54+5n$ instead of $-54+5(n-1)$ ; that forgets the crucial $(n-1)$ shift. The formula $5n-49$ is off by $+10$ : substituting $n=9$ gives $-4$ , not $-14$ . The formula $5n-64$ is off by $-5$ : substituting $n=9$ gives $-19$ . A common source of that last error is finding $a_1=26-17(5)=-59$ by using $17$ steps back from the 17th term, rather than $16$ steps. That off-by-one mistake creates an intercept that is $5$ too small. So when you solve this type of problem, always compute $d$ first, then find $a_1$ by stepping back with $(\text{term index}-1)d$ . Finally, test your formula by plugging in one of the given term numbers—that catches intercept errors quickly.
The sum of the first $n$ terms of an arithmetic series is given by $S_n = 3n^2 - n$ . What is the common difference?
Answer: $6$ Whenever you see the sum of the first $n$ terms of an arithmetic series written as a quadratic like $S_n = ̂3n^2 - n$ , think about the general form. For an arithmetic series, $S_n = \frac n2\left[2a + (n-1)d\right] = \frac d2 n^2 + \left(a - \frac d2\right)n.$ So the coefficient of $n^2$ is $\frac d2$ , not simply $d$ . Here the coefficient is $3$ , so $\frac d2 = ̂3$ , which gives $d=6$ . That is why the correct answer is $6$ . The distractor $3$ is just the coefficient of $n^2$ ; students often mistake it for the common difference, but it is only half of $d$ . The choice $2$ might come from confusing the coefficient of $n$ with the difference, or from dividing $6$ by something incorrectly;it has no direct role in the sum formula's leading coefficient. The choice $8$ could result from an arithmetic slip like $3\cdot 2 + ̂2$ , but there is no support for it in the comparison. For your study tip: on this exam, whenever an arithmetic series sum is given as $S_n = an^2 + bn$ , the common difference is always $2a$ . Remember the factor of $2$ comes from the formula's $\frac d2 n^2$ . Then check the linear coefficient if needed to find the first term, but the difference is determined entirely by the quadratic coefficient.
An auditorium has 20 rows of seats. Row 1 contains 20 seats, row 9 contains 44 seats, and the number of seats in consecutive rows increases by a constant amount. What is the total number of seats in the auditorium?
Answer: $970$ Whenever you see consecutive quantities increasing by a constant amount, you are working with an arithmetic sequence. Here the row counts are the terms, so find the common difference first. Row 1 has 20 seats and row 9 has 44 seats; from row 1 to row 9 is 8 steps, so each step is $(44-20)/8 = 3$ seats per row. Thus row 20 has $20 + 19(3) = 77$ seats. Since an arithmetic sequence sum depends only on the first and last terms, the total is $\frac{20}{2}(20+77) = 10(97) = 970$ . The choice $1000$ comes from a common trap: using 20 as the number of increases, so row 20 becomes $20 + 20(3) = 80$ , then averaging 20 and 80. But there are only 19 increases between row 1 and row 20. The choice $1940$ uses the correct 77 but forgets to divide by 2: $20(20+77)$ gives twice the sum. The choice $2000$ combines both mistakes: it uses row 20 = 80 and also fails to divide by 2, giving $20(20+80)$ . On exam day, write out the pattern and check the number of intervals: from row 1 to row $n$ is $n-1$ increases. Then apply the sum formula, and don't forget the factor $\frac{n}{2}$ .
The sum of the first $n$ terms of an arithmetic series is $S_n=3n^2-2n$ . Which expression gives $a_n$ , the $n$ th term?
Answer: $6n-5$ Since $a_n=S_n-S_{n-1}$ , compute $S_{n-1}=3(n-1)^2-2(n-1)=3n^2-8n+5$ . Subtracting gives $a_n=(3n^2-2n)-(3n^2-8n+5)=6n-5$ . The expression $3n-2$ is $S_n/n$ , $3n^2-2n$ is the sum itself, and $6n+1$ is $a_{n+1}$ from $S_{n+1}-S_n$ .
Five arithmetic means are inserted between $8$ and $44$ . What is the sum of the five inserted means?
Answer: $130$ Whenever you insert arithmetic means, think of them as terms of one arithmetic sequence: the endpoints are the first and last terms, and each inserted mean is a term in between. Here five means are inserted, so the sequence has $7$ terms total. The common difference is $d=\frac{44-8}{5+1}=6.$ So the five means are $14,20,26,32,38$ , and their sum is $14+20+26+32+38=130.$ Notice that the sum of all seven terms is $\frac{8+44}{2}\cdot 7=182$ . That explains the trap behind 182: it is the total of everything, including both endpoints. The choice 156 would result from mistakenly inserting six means instead of five, since six means would have sum $6\cdot 26=156$ . The choice 174 equals $14+20+26+32+38+44$ , so it includes the last endpoint but still leaves out the first endpoint $8$ . When you see "arithmetic means," set up the sequence immediately: first term is the lower endpoint, last term is the upper endpoint, and the number of terms is the number of means plus $2$ . Then use $d=\frac{\text{last}-\text{first}}{\text{number of means}+1}$ and list the means. This protects you from both endpoint and interval-counting mistakes.
In an arithmetic sequence, the 3rd term is $19$ and the 8th term is $4$ . What is the sum of all positive terms?
Answer: $117$ Whenever you see an arithmetic sequence question, your first move is to find the common difference. Here, the 3rd term is $19$ and the 8th term is $4$ . Since there are $5$ steps from term 3 to term 8, the difference is $\frac{4-19}{5}=-3$ . Working backward, the first term is $19 - 2(-3)=25$ , so the sequence is $25, 22, 19, \dots$ . A term is positive as long as $25+(n-1)(-3)>0$ , which gives $n<9\tfrac{1}{3}$ , so the positive terms are terms 1 through 9. The 9th term is $25+8(-3)=1$ . The sum of the first 9 terms is $\frac{9}{2}(25+1)=117$ . The other choices come from stopping at the wrong place. $116$ is the sum of the first 8 terms, so it leaves out the 9th term, which is still positive $1$ . $115$ is the sum of the first 10 terms, but the 10th term is $-2$ , and negative terms should not be included. $112$ is the sum of the first 7 terms, stopping before the positive 8th term $4$ and 9th term $1$ are added. For any arithmetic sequence sum question, first determine the common difference, then carefully identify exactly which terms satisfy the condition—here, being positive. Counting one term too few or too many is the most common trap.
Seven arithmetic means are inserted between $-13$ and $43$ . What is the common difference of the resulting sequence?
Answer: 7 The resulting sequence has $7+2=9$ terms, hence $8$ gaps. Therefore $d=\frac{43-(-13)}{8}=\frac{56}{8}=7$ . The value $\frac{56}{9}$ uses the number of terms as the number of gaps; $8$ comes from dividing by the 7 inserted means; $\frac{28}{3}$ uses 6 gaps.

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