Sequences and series

Answer: $-43$

The common difference is $d = 2 - 11 = -9\mathclose{.}$ The $n$th term is $a + (n - 1)d$ — the step is taken $6$ times, not $7\mathclose{:}$ $11 + 6 \times (-9) = -43\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $a + nd\mathclose{,}$ taking the step one time too many → $-52$
• Left out the first term, giving $(n - 1)d$ → $-54$

Answer: $259$

The last term is $10 + 6 \times 9 = 64\mathclose{,}$ and an arithmetic series adds up to the number of terms times the mean of the first and last: $7 \times \dfrac{10 + 64}{2} = 259\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the number of terms by the LAST term, not the average → $448$
• Multiplied the number of terms by the FIRST term → $70$

Answer: $-10$

The ratio is $r = -2$ and the first term $a = 2\mathclose{.}$ The sum of $n$ terms is $\dfrac{a(r^n - 1)}{r - 1} = \dfrac{2\left((-2)^{4} - 1\right)}{-2 - 1} = -10\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $r^{n-1}$ instead of $r^n$ in the formula → $6$
• Gave the 4th term instead of the sum → $-16$

Answer: $-128$

The common ratio is $r = -8 \div 4 = -2\mathclose{.}$ The $n$th term is $ar^{n-1}\mathclose{,}$ so the ratio is applied $5$ times: $4 \times \left(-2\right)^{5} = -128\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $ar^n\mathclose{,}$ applying the ratio one time too many → $256$
• Added the ratio each time instead of multiplying → $-6$

Answer: $20$

The common difference is $d = -8\mathclose{,}$ and term number $n$ is $9 + (n - 1) \times (-8)\mathclose{.}$ Setting that equal to $-143\mathclose{:}$ $(n - 1) \times (-8) = -152\mathclose{,}$ so $n - 1 = 19$ and $n = 20\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to add 1 at the end → $19$

Answer: $160$

The terms run from $k = 1\mathclose{,}$ giving $-1\mathclose{,}$ to $k = 8\mathclose{,}$ giving $41\mathclose{,}$ going up by $6$ each time — an arithmetic series of $8$ terms. Its sum is $8 \times \dfrac{-1 + 41}{2} = 160\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the constant, adding up only $6k$ → $216$
• Added the constant only once instead of once per term → $209$

Answer: $\dfrac{99}{4}$

The ratio is $r = -\dfrac{1}{3}\mathclose{,}$ and $|r| < 1\mathclose{,}$ so the series has a sum: $\dfrac{a}{1 - r} = \dfrac{33}{1 - \left(-\dfrac{1}{3}\right)} = \dfrac{99}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $1 + r$ in the denominator → $\frac{99}{2}$
• Multiplied by $1 - r$ instead of dividing by it → $44$

Answer: $123$

The common difference is $d = 11 - 7 = 4\mathclose{.}$ The $n$th term is $a + (n - 1)d$ — the step is taken $29$ times, not $30\mathclose{:}$ $7 + 29 \times 4 = 123\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $a + nd\mathclose{,}$ taking the step one time too many → $127$
• Left out the first term, giving $(n - 1)d$ → $116$

Answer: $\dfrac{3}{2}$

The common ratio is $r = 24 \div 48 = \dfrac{1}{2}\mathclose{.}$ The $n$th term is $ar^{n-1}\mathclose{,}$ so the ratio is applied $5$ times: $48 \times \left(\frac{1}{2}\right)^{5} = \dfrac{3}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $ar^n\mathclose{,}$ applying the ratio one time too many → $\frac{3}{4}$
• Added the ratio each time instead of multiplying → $\frac{101}{2}$

Answer: $84$

The last term is $-3 + 6 \times 5 = 27\mathclose{,}$ and an arithmetic series adds up to the number of terms times the mean of the first and last: $7 \times \dfrac{-3 + 27}{2} = 84\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the number of terms by the LAST term, not the average → $189$
• Multiplied the number of terms by the FIRST term → $-21$

Answer: $364$

The ratio is $r = 3$ and the first term $a = 1\mathclose{.}$ The sum of $n$ terms is $\dfrac{a(r^n - 1)}{r - 1} = \dfrac{3^{6} - 1}{3 - 1} = 364\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $r^{n-1}$ instead of $r^n$ in the formula → $121$
• Gave the 6th term instead of the sum → $243$

Answer: $16$

The common difference is $d = 4\mathclose{,}$ and term number $n$ is $-3 + (n - 1) \times 4\mathclose{.}$ Setting that equal to $57\mathclose{:}$ $(n - 1) \times 4 = 60\mathclose{,}$ so $n - 1 = 15$ and $n = 16\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to add 1 at the end → $15$

Answer: $135$

The terms run from $k = 1\mathclose{,}$ giving $0\mathclose{,}$ to $k = 10\mathclose{,}$ giving $27\mathclose{,}$ going up by $3$ each time — an arithmetic series of $10$ terms. Its sum is $10 \times \dfrac{0 + 27}{2} = 135\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the constant, adding up only $3k$ → $165$
• Added the constant only once instead of once per term → $162$

Answer: $32$

The ratio is $r = \dfrac{1}{4}\mathclose{,}$ and $|r| < 1\mathclose{,}$ so the series has a sum: $\dfrac{a}{1 - r} = \dfrac{24}{1 - \dfrac{1}{4}} = 32\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $1 + r$ in the denominator → $\frac{96}{5}$
• Multiplied by $1 - r$ instead of dividing by it → $18$

Answer: $74$

The common difference is $d = 9 - 4 = 5\mathclose{.}$ The $n$th term is $a + (n - 1)d$ — the step is taken $14$ times, not $15\mathclose{:}$ $4 + 14 \times 5 = 74\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $a + nd\mathclose{,}$ taking the step one time too many → $79$
• Left out the first term, giving $(n - 1)d$ → $70$

Answer: $9$

The common ratio is $r = 36 \div 72 = \dfrac{1}{2}\mathclose{.}$ The $n$th term is $ar^{n-1}\mathclose{,}$ so the ratio is applied $3$ times: $72 \times \left(\frac{1}{2}\right)^{3} = 9\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $ar^n\mathclose{,}$ applying the ratio one time too many → $\frac{9}{2}$
• Added the ratio each time instead of multiplying → $\frac{147}{2}$

Answer: $2772$

The last term is $12 + 23 \times 9 = 219\mathclose{,}$ and an arithmetic series adds up to the number of terms times the mean of the first and last: $24 \times \dfrac{12 + 219}{2} = 2772\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the number of terms by the LAST term, not the average → $5256$
• Multiplied the number of terms by the FIRST term → $288$

Answer: $-25$

The ratio is $r = -2$ and the first term $a = 5\mathclose{.}$ The sum of $n$ terms is $\dfrac{a(r^n - 1)}{r - 1} = \dfrac{5\left((-2)^{4} - 1\right)}{-2 - 1} = -25\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $r^{n-1}$ instead of $r^n$ in the formula → $15$
• Gave the 4th term instead of the sum → $-40$

Answer: $23$

The common difference is $d = 2\mathclose{,}$ and term number $n$ is $-11 + (n - 1) \times 2\mathclose{.}$ Setting that equal to $33\mathclose{:}$ $(n - 1) \times 2 = 44\mathclose{,}$ so $n - 1 = 22$ and $n = 23\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to add 1 at the end → $22$

Answer: $286$

The terms run from $k = 1\mathclose{,}$ giving $4\mathclose{,}$ to $k = 13\mathclose{,}$ giving $40\mathclose{,}$ going up by $3$ each time — an arithmetic series of $13$ terms. Its sum is $13 \times \dfrac{4 + 40}{2} = 286\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the constant, adding up only $3k$ → $273$
• Added the constant only once instead of once per term → $274$

Answer: $\dfrac{128}{3}$

The ratio is $r = \dfrac{1}{4}\mathclose{,}$ and $|r| < 1\mathclose{,}$ so the series has a sum: $\dfrac{a}{1 - r} = \dfrac{32}{1 - \dfrac{1}{4}} = \dfrac{128}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $1 + r$ in the denominator → $\frac{128}{5}$
• Multiplied by $1 - r$ instead of dividing by it → $24$

Answer: $162$

The common ratio is $r = 6 \div 2 = 3\mathclose{.}$ The $n$th term is $ar^{n-1}\mathclose{,}$ so the ratio is applied $4$ times: $2 \times 3^{4} = 162\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $ar^n\mathclose{,}$ applying the ratio one time too many → $486$
• Added the ratio each time instead of multiplying → $14$

Answer: $78$

The common difference is $d = 8 - 1 = 7\mathclose{.}$ The $n$th term is $a + (n - 1)d$ — the step is taken $11$ times, not $12\mathclose{:}$ $1 + 11 \times 7 = 78\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $a + nd\mathclose{,}$ taking the step one time too many → $85$
• Left out the first term, giving $(n - 1)d$ → $77$

Answer: $150$

The last term is $-6 + 24 \times 1 = 18\mathclose{,}$ and an arithmetic series adds up to the number of terms times the mean of the first and last: $25 \times \dfrac{-6 + 18}{2} = 150\mathclose{.}$

Common mistakes, and the answer each one gives:

• Multiplied the number of terms by the LAST term, not the average → $450$
• Multiplied the number of terms by the FIRST term → $-150$

Answer: $121$

The ratio is $r = 3$ and the first term $a = 1\mathclose{.}$ The sum of $n$ terms is $\dfrac{a(r^n - 1)}{r - 1} = \dfrac{3^{5} - 1}{3 - 1} = 121\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $r^{n-1}$ instead of $r^n$ in the formula → $40$
• Gave the 5th term instead of the sum → $81$

Answer: $23$

The common difference is $d = 7\mathclose{,}$ and term number $n$ is $-4 + (n - 1) \times 7\mathclose{.}$ Setting that equal to $150\mathclose{:}$ $(n - 1) \times 7 = 154\mathclose{,}$ so $n - 1 = 22$ and $n = 23\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to add 1 at the end → $22$

Answer: $32$

The common ratio is $r = 4 \div 2 = 2\mathclose{.}$ The $n$th term is $ar^{n-1}\mathclose{,}$ so the ratio is applied $4$ times: $2 \times 2^{4} = 32\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $ar^n\mathclose{,}$ applying the ratio one time too many → $64$
• Added the ratio each time instead of multiplying → $10$

Answer: $1116$

The terms run from $k = 1\mathclose{,}$ giving $11\mathclose{,}$ to $k = 18\mathclose{,}$ giving $113\mathclose{,}$ going up by $6$ each time — an arithmetic series of $18$ terms. Its sum is $18 \times \dfrac{11 + 113}{2} = 1116\mathclose{.}$

Common mistakes, and the answer each one gives:

• Left out the constant, adding up only $6k$ → $1026$
• Added the constant only once instead of once per term → $1031$

Answer: $36$

The ratio is $r = \dfrac{2}{3}\mathclose{,}$ and $|r| < 1\mathclose{,}$ so the series has a sum: $\dfrac{a}{1 - r} = \dfrac{12}{1 - \dfrac{2}{3}} = 36\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $1 + r$ in the denominator → $\frac{36}{5}$
• Multiplied by $1 - r$ instead of dividing by it → $4$

Answer: $44$

The common difference is $d = 11 - 8 = 3\mathclose{.}$ The $n$th term is $a + (n - 1)d$ — the step is taken $12$ times, not $13\mathclose{:}$ $8 + 12 \times 3 = 44\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $a + nd\mathclose{,}$ taking the step one time too many → $47$
• Left out the first term, giving $(n - 1)d$ → $36$