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Neokaal Study

Arithmetic Progressions

Competency Practice

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

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  1. 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.

  2. 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 $0$.

  3. Which term of $15, 10, 5, 0, \ldots$ is the first negative term?

  4. A sequence has $a_n=6n+5$. Show that it is an AP and find $S_{8}$.

  5. An AP has $a_{2}=4$ and $a_{6}=24$. Find $a_{13}$.

  6. Is $8$ a term of the AP $-4, -2, 0, 2, \ldots$? If yes, state its position.

  7. Find the sum of the first 18 terms of $-2, 3, 8, 13, \ldots$.

  8. An AP has first term $9$, last term $27$ and sum $180$. How many terms does it contain?

Neokaal Study

Answer Key

  1. (i) $536$ units, (ii) $1211$ units, (iii) $5327$ units

    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.

  2. $S_{m + n} = 0$

    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).

  3. The 5th term, equal to $-5$.

    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 $a_{5}=-5$.

  4. It is an AP with $a=11$, $d=6$, and $S_{8}=256$.

    1. $a_{n+1}-a_n=6$, which is constant.
    2. Thus $a=11$ and $d=6$.
    3. The sum formula gives $S_{8}=256$.

  5. $a_{13}=59$

    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 $a_{13}=a+(13-1)d=59$.

  6. Yes, it is term 7.

    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.

  7. $729$

    1. Read $a=-2$ and $d=5$ from the sequence.
    2. Apply $S_n=\frac{n}{2}[2a+(n-1)d]$.
    3. After substitution, $S_{18}=729$.

  8. $10$ terms

    1. For a finite AP, $S_n=\frac{n}{2}(a+l)$.
    2. So $180=\frac{n}{2}(9+27)$.
    3. Solving gives $n=10$.