NeokaalStudy

Arithmetic Progressions: Practice Worksheet

Terms, common differences, finite sums, sequence patterns, and applications.

18 problems·35–45 min·★★★★★
Print worksheet (PDF) →
  1. Problem 1
    Identifying Arithmetic Progressions·★★☆☆☆
    Does the list $5, 9, 13, 17, \ldots$ begin an arithmetic progression? Justify your answer.
    ▶Answer
    Yes.
    ▶Step-by-step solution
    1. The consecutive differences are 4, 4, 4.
    2. They are equal, so the common difference is fixed.
    3. Therefore the list is an AP.
    Report an issue →
  2. Problem 2
    Constructing Arithmetic Progressions·★★☆☆☆
    Write the first four terms of the AP with $a=2$ and $d=4$.
    ▶Answer
    $2, 6, 10, 14$
    ▶Step-by-step solution
    1. Start with the first term $a$.
    2. Add the common difference $4$ to obtain each new term.
    3. The terms are $2, 6, 10, 14$.
    Report an issue →
  3. Problem 3
    Identifying Arithmetic Progressions·★★☆☆☆
    Verify that $-6, -3, 0, 3$ are consecutive terms of an AP, then write the next three terms.
    ▶Answer
    The next terms are $6, 9, 12$.
    ▶Step-by-step solution
    1. Each consecutive difference equals $3$.
    2. Hence the list is an arithmetic progression.
    3. Add $3$ repeatedly to obtain $6, 9, 12$.
    Report an issue →
  4. Problem 4
    nth Term of an AP·★★★☆☆
    Find the 10th term of the AP $2, 7, 12, 17, \ldots$.
    ▶Answer
    $47$
    ▶Step-by-step solution
    1. The first term is $a=2$ and the common difference is $d=5$.
    2. Use $a_n=a+(n-1)d$ with $n=10$.
    3. So $a_{10}=2+(10-1)(5)=47$.
    Report an issue →
  5. Problem 5
    Term Position·★★★☆☆
    Which term of the AP $-2, -1, 0, 1, \ldots$ is $12$?
    ▶Answer
    The 15th term.
    ▶Step-by-step solution
    1. Write $a_n=a+(n-1)d$.
    2. Substitute $a=-2$, $d=1$ and $a_n=12$.
    3. Solving gives $n=15$, so the value occurs at term 15.
    Report an issue →
  6. Problem 6
    Common Difference·★★★☆☆
    An AP has common difference $-3$. Find $a_{10}-a_{4}$.
    ▶Answer
    $-18$
    ▶Step-by-step solution
    1. For an AP, $a_q-a_p=(q-p)d$.
    2. Here $q-p=6$ and $d=-3$.
    3. Therefore $a_{10}-a_{4}=(6)(-3)=-18$.
    Report an issue →
  7. Problem 7
    Common Difference·★★★★☆
    In an AP, $a_{12}-a_{7}=15$. Find the common difference.
    ▶Answer
    $d=3$
    ▶Step-by-step solution
    1. Use $a_q-a_p=(q-p)d$.
    2. Thus $(12-7)d=15$.
    3. Dividing by 5 gives $d=3$.
    Report an issue →
  8. Problem 8
    nth Term of an AP·★★★★☆
    An AP has $a_{3}=2$ and $a_{5}=8$. Find $a_{9}$.
    ▶Answer
    $a_{9}=20$
    ▶Step-by-step solution
    1. Subtract the term equations: $(5-3)d=6$, so $d=3$.
    2. Use $a_{3}=a+(3-1)d$ to obtain $a=-4$.
    3. Then $a_{9}=a+(9-1)d=20$.
    Report an issue →
  9. Problem 9
    Term Position·★★★★☆
    Is $22$ a term of the AP $-2, 4, 10, 16, \ldots$? If yes, state its position.
    ▶Answer
    Yes, it is term 5.
    ▶Step-by-step solution
    1. Set $a_n=22$ in $a_n=a+(n-1)d$.
    2. This gives $n=5$.
    3. The value is a positive integer, so it is a valid term position.
    Report an issue →
  10. Problem 10
    Term from the End·★★★★☆
    Find the 7th term from the end of the finite AP $4, 6, 8, \ldots$, whose last term is $28$.
    ▶Answer
    $16$
    ▶Step-by-step solution
    1. The last term is $a_{13}$, so the AP has 13 terms.
    2. The 7th term from the end is term $13-7+1=7$.
    3. Hence the required value is $a_{7}=16$.
    Report an issue →
  11. Problem 11
    Determining an Arithmetic Progression·★★★★★
    The 6th term of an AP is $-29$, and $a_{16}-a_{11}=-25$. Determine its first term and common difference.
    ▶Answer
    $a=-4$, $d=-5$
    ▶Step-by-step solution
    1. Since $a_{16}-a_{11}=(5)d$, $d=-5$.
    2. Use $a_{6}=a+(6-1)d=-29$.
    3. Solving gives $a=-4$, so the AP is $-4, -9, -14, -19, \ldots$.
    Report an issue →
  12. Problem 12
    Term Position·★★★★★
    Which term of $8, 4, 0, -4, \ldots$ is the first negative term?
    ▶Answer
    The 4th term, equal to $-4$.
    ▶Step-by-step solution
    1. Solve $a+(n-1)d<0$ for the least positive integer $n$.
    2. The first integer satisfying the inequality is $n=4$.
    3. Indeed $a_{3}=0$ and $a_{4}=-4$.
    Report an issue →
  13. Problem 13
    nth Term and Sum·★★★★★
    A sequence has $a_n=3n+6$. Show that it is an AP and find $S_{20}$.
    ▶Answer
    It is an AP with $a=9$, $d=3$, and $S_{20}=750$.
    ▶Step-by-step solution
    1. $a_{n+1}-a_n=3$, which is constant.
    2. Thus $a=9$ and $d=3$.
    3. The sum formula gives $S_{20}=750$.
    Report an issue →
  14. Problem 14
    Term Position and Sum·★★★★★
    For the AP $3, -2, -7, -12, \ldots$, determine the position of $-42$ and find the sum through that term.
    ▶Answer
    It is term 10, and the sum is $-195$.
    ▶Step-by-step solution
    1. Solve $a+(n-1)d=-42$ to get $n=10$.
    2. Then use the AP sum formula for the first $n$ terms.
    3. This gives $S_{10}=-195$.
    Report an issue →
  15. Problem 15
    Sum of an AP·★★☆☆☆
    An AP starts with $2$ and has common difference $-3$. Find the sum of its first 7 terms.
    ▶Answer
    $S_{7}=-49$
    ▶Step-by-step solution
    1. Use $S_n=\frac{n}{2}[2a+(n-1)d]$.
    2. Substitute $n=7$, $a=2$ and $d=-3$.
    3. This gives $S_{7}=-49$.
    Report an issue →
  16. Problem 16
    Sum of an AP·★★☆☆☆
    Find the sum of the first 10 terms of $-2, 0, 2, 4, \ldots$.
    ▶Answer
    $70$
    ▶Step-by-step solution
    1. Read $a=-2$ and $d=2$ from the sequence.
    2. Apply $S_n=\frac{n}{2}[2a+(n-1)d]$.
    3. After substitution, $S_{10}=70$.
    Report an issue →
  17. Problem 17
    Number of Terms·★★★☆☆
    An AP has first term $8$, last term $40$ and sum $408$. How many terms does it contain?
    ▶Answer
    $17$ terms
    ▶Step-by-step solution
    1. For a finite AP, $S_n=\frac{n}{2}(a+l)$.
    2. So $408=\frac{n}{2}(8+40)$.
    3. Solving gives $n=17$.
    Report an issue →
  18. Problem 18
    Sum of an AP·★★★☆☆
    The finite AP $4, 9, 14, \ldots$ ends at $94$. Find the sum of its last 5 terms.
    ▶Answer
    $420$
    ▶Step-by-step solution
    1. The final value is term 19.
    2. The last 5 terms begin at term 15, equal to $74$.
    3. Their AP sum is $420$.
    Report an issue →