Practice Worksheet

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

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Does the list $5, 9, 13, 17, \ldots$ begin an arithmetic progression? Justify your answer.
Write the first four terms of the AP with $a=2$ and $d=4$.
Verify that $-6, -3, 0, 3$ are consecutive terms of an AP, then write the next three terms.
Find the 10th term of the AP $2, 7, 12, 17, \ldots$.
Which term of the AP $-2, -1, 0, 1, \ldots$ is $12$?
An AP has common difference $-3$. Find $a_{10}-a_{4}$.
In an AP, $a_{12}-a_{7}=15$. Find the common difference.
An AP has $a_{3}=2$ and $a_{5}=8$. Find $a_{9}$.
Is $22$ a term of the AP $-2, 4, 10, 16, \ldots$? If yes, state its position.
Find the 7th term from the end of the finite AP $4, 6, 8, \ldots$, whose last term is $28$.
The 6th term of an AP is $-29$, and $a_{16}-a_{11}=-25$. Determine its first term and common difference.
Which term of $8, 4, 0, -4, \ldots$ is the first negative term?
A sequence has $a_n=3n+6$. Show that it is an AP and find $S_{20}$.
For the AP $3, -2, -7, -12, \ldots$, determine the position of $-42$ and find the sum through that term.
An AP starts with $2$ and has common difference $-3$. Find the sum of its first 7 terms.
Find the sum of the first 10 terms of $-2, 0, 2, 4, \ldots$.
An AP has first term $8$, last term $40$ and sum $408$. How many terms does it contain?
The finite AP $4, 9, 14, \ldots$ ends at $94$. Find the sum of its last 5 terms.

Neokaal Study

Answer Key

Yes.
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.
$2, 6, 10, 14$
1. Start with the first term $a$.
2. Add the common difference $4$ to obtain each new term.
3. The terms are $2, 6, 10, 14$.
The next terms are $6, 9, 12$.
1. Each consecutive difference equals $3$.
2. Hence the list is an arithmetic progression.
3. Add $3$ repeatedly to obtain $6, 9, 12$.
$47$
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 $a_{10}=2+(10-1)(5)=47$.
The 15th term.
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.
$-18$
1. For an AP, $a_q-a_p=(q-p)d$.
2. Here $q-p=6$ and $d=-3$.
3. Therefore $a_{10}-a_{4}=(6)(-3)=-18$.
$d=3$
1. Use $a_q-a_p=(q-p)d$.
2. Thus $(12-7)d=15$.
3. Dividing by 5 gives $d=3$.
$a_{9}=20$
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 $a_{9}=a+(9-1)d=20$.
Yes, it is term 5.
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.
$16$
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 $a_{7}=16$.
$a=-4$, $d=-5$
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 $-4, -9, -14, -19, \ldots$.
The 4th term, equal to $-4$.
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 $a_{4}=-4$.
It is an AP with $a=9$, $d=3$, and $S_{20}=750$.
1. $a_{n+1}-a_n=3$, which is constant.
2. Thus $a=9$ and $d=3$.
3. The sum formula gives $S_{20}=750$.
It is term 10, and the sum is $-195$.
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 $S_{10}=-195$.
$S_{7}=-49$
1. Use $S_n=\frac{n}{2}[2a+(n-1)d]$.
2. Substitute $n=7$, $a=2$ and $d=-3$.
3. This gives $S_{7}=-49$.
$70$
1. Read $a=-2$ and $d=2$ from the sequence.
2. Apply $S_n=\frac{n}{2}[2a+(n-1)d]$.
3. After substitution, $S_{10}=70$.
$17$ terms
1. For a finite AP, $S_n=\frac{n}{2}(a+l)$.
2. So $408=\frac{n}{2}(8+40)$.
3. Solving gives $n=17$.
$420$
1. The final value is term 19.
2. The last 5 terms begin at term 15, equal to $74$.
3. Their AP sum is $420$.