Arithmetic Series: Additional Practice (Set 1)

The summand $4k - 3$ gives arithmetic terms with $a_1 = 1$ and $d = 4$, so $a_{12} = 4 \times 12 - 3 = 45$.

$S_{12} = \frac{12(1 + 45)}{2} = 6 \times 46 = 276$

From $\frac{n(n+1)}{2} = 325$ we get $n(n+1) = 650$. Look for consecutive factors of $650$.

$n(n+1) = 25 \times 26 = 650 \;\Rightarrow\; n = 25$

The index runs from $5$ to $15$, a count of $15 - 5 + 1 = 11$, with first term $5$ and last term $15$.

$S = \frac{11(5 + 15)}{2} = 11 \times 10 = 110$

The twentieth term is $a_{20} = 100 + 19 \times (-4) = 24$, keeping the negative $d$ through the step.

$S_{20} = \frac{20(100 + 24)}{2} = 10 \times 124 = 1240$

There are $50$ odd numbers from $1$ to $99$, so $n = 50$, $a_1 = 1$, and $a_{50} = 99$.

$S_{50} = \frac{50(1 + 99)}{2} = 25 \times 100 = 2500$

Use $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, so $185 = \frac{10}{2}(10 + 9d) = 5(10 + 9d)$, giving $10 + 9d = 37$.

$9d = 27 \;\Rightarrow\; d = 3$

The summand $3k + 2$ gives arithmetic terms with $a_1 = 5$ and $d = 3$, so $a_8 = 3 \times 8 + 2 = 26$.

$S_8 = \frac{8(5 + 26)}{2} = 4 \times 31 = 124$

With $n = 30$, $a_1 = 2$, and $a_{30} = 60$, apply the series formula (which matches $n(n+1) = 30 \times 31$).

$S_{30} = \frac{30(2 + 60)}{2} = 15 \times 62 = 930$

The seats form an arithmetic sequence with $a_1 = 22$ and $d = 2$, so the last row is $a_{30} = 22 + 29 \times 2 = 80$.

$S_{30} = \frac{30(22 + 80)}{2} = 15 \times 102 = 1530$

The sum of the first $n$ odd numbers is $n^2$, and here $n = 15$; you can also add directly from $a_1 = 1$ to $a_{15} = 29$.

$S_{15} = \frac{15(1 + 29)}{2} = 15 \times 15 = 225$

From $a_8 - a_3 = 5d = 15$ we get $d = 3$, and $a_1 = a_3 - 2d = 12 - 6 = 6$.

Now sum $8$ terms with first term $6$ and last term $27$.

$S_8 = \frac{8(6 + 27)}{2} = 4 \times 33 = 132$

Set $\frac{n}{2}(3n + 11) = 585$, so $n(3n + 11) = 1170$, giving $3n^2 + 11n - 1170 = 0$.

The positive whole-number root is

$n = \frac{-11 + \sqrt{14161}}{6} = \frac{-11 + 119}{6} = 18$