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Arithmetic Progressions: Competency Practice

Industrial production schedules, algebraic sum proofs, first negative term thresholds, and series modeling.

8 problems·20–25 min·★★★★★
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  1. Problem 1
    arithmetic-progressions·★★★★★
    A green mobility manufacturer increases its production of electric scooters uniformly by a fixed number of units every year. The factory produced $686$ units in the $3^{\text{rd}}$ year and $986$ units in the $7^{\text{th}}$ year. (i) Find the production in the first year. (ii) Find the production in the $10^{\text{th}}$ year. (iii) Find the total production in the first $7$ years.
    ▶Answer
    (i) $536$ units, (ii) $1211$ units, (iii) $5327$ units
    ▶Step-by-step solution
    1. Since annual production increases uniformly, the yearly outputs form an AP with first term $a$ and common difference $d$.
    2. Given $a_{3} = a + 2d = 686$ and $a_{7} = a + 6d = 986$.
    3. Subtracting the equations gives $4d = 300 \implies d = 75$ units per year.
    4. (i) First year production: $a = 686 - 2(75) = 536$ units.
    5. (ii) Production in year $10$: $a_{10} = 536 + (10 - 1)(75) = 1211$ units.
    6. (iii) Total production in first $7$ years: $S_{7} = \frac{7}{2}[2(536) + (7 - 1)(75)] = 5327$ units.
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  2. Problem 2
    arithmetic-progressions·★★★★★
    If the sum of the first $m$ terms of an Arithmetic Progression is equal to the sum of its first $n$ terms ($m \ne n$), show that the sum of its first $(m + n)$ terms is equal to $0$.
    ▶Answer
    $S_{m + n} = 0$
    ▶Step-by-step solution
    1. Let the first term of the AP be $a$ and the common difference be $d$. The sum formula is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
    2. Given $S_{m} = S_{n}: \frac{m}{2}[2a + (m - 1)d] = \frac{n}{2}[2a + (n - 1)d]$.
    3. Expanding and collecting terms: $2a(m - n) + [(m^2 - n^2) - (m - n)]d = 0$.
    4. Factoring $(m - n)$ gives $(m - n)[2a + (m + n - 1)d] = 0$. Since $m \ne n$, $2a + (m + n - 1)d = 0$.
    5. Therefore, $S_{m + n} = \frac{m + n}{2}[2a + (m + n - 1)d] = \frac{m + n}{2} \times (0) = 0$ (Hence proved).
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  3. Problem 3
    Term Position·★★★★☆
    Which term of $15, 10, 5, 0, \ldots$ is the first negative term?
    ▶Answer
    The 5th term, equal to $-5$.
    ▶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 $n=5$.
    3. Indeed $a_{4}=0$ and $a_{5}=-5$.
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  4. Problem 4
    nth Term and Sum·★★★☆☆
    A sequence has $a_n=6n+5$. Show that it is an AP and find $S_{8}$.
    ▶Answer
    It is an AP with $a=11$, $d=6$, and $S_{8}=256$.
    ▶Step-by-step solution
    1. $a_{n+1}-a_n=6$, which is constant.
    2. Thus $a=11$ and $d=6$.
    3. The sum formula gives $S_{8}=256$.
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  5. Problem 5
    nth Term of an AP·★★★★☆
    An AP has $a_{2}=4$ and $a_{6}=24$. Find $a_{13}$.
    ▶Answer
    $a_{13}=59$
    ▶Step-by-step solution
    1. Subtract the term equations: $(6-2)d=20$, so $d=5$.
    2. Use $a_{2}=a+(2-1)d$ to obtain $a=-1$.
    3. Then $a_{13}=a+(13-1)d=59$.
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  6. Problem 6
    Term Position·★★★☆☆
    Is $8$ a term of the AP $-4, -2, 0, 2, \ldots$? If yes, state its position.
    ▶Answer
    Yes, it is term 7.
    ▶Step-by-step solution
    1. Set $a_n=8$ in $a_n=a+(n-1)d$.
    2. This gives $n=7$.
    3. The value is a positive integer, so it is a valid term position.
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  7. Problem 7
    Sum of an AP·★★★★☆
    Find the sum of the first 18 terms of $-2, 3, 8, 13, \ldots$.
    ▶Answer
    $729$
    ▶Step-by-step solution
    1. Read $a=-2$ and $d=5$ from the sequence.
    2. Apply $S_n=\frac{n}{2}[2a+(n-1)d]$.
    3. After substitution, $S_{18}=729$.
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  8. Problem 8
    Number of Terms·★★★★☆
    An AP has first term $9$, last term $27$ and sum $180$. How many terms does it contain?
    ▶Answer
    $10$ terms
    ▶Step-by-step solution
    1. For a finite AP, $S_n=\frac{n}{2}(a+l)$.
    2. So $180=\frac{n}{2}(9+27)$.
    3. Solving gives $n=10$.
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