Arithmetic Series: Additional Practice (Set 2)

Start from $S_n = \frac{n(a_1 + a_n)}{2}$, so $306 = \frac{12(3 + a_{12})}{2} = 6(3 + a_{12})$.

Divide by $6$ and solve for the last term.

$3 + a_{12} = 51 \;\Rightarrow\; a_{12} = 48$

Here $a_1 = 3$ and $d = 4$, so the fortieth term is $a_{40} = 3 + 39 \times 4 = 159$.

$S_{40} = \frac{40(3 + 159)}{2} = 20 \times 162 = 3240$

Take the whole sum up to $100$ and remove the sum up to $9$: $\frac{100 \times 101}{2} - \frac{9 \times 10}{2} = 5050 - 45$.

$10 + 11 + \cdots + 100 = 5050 - 45 = 5005$

With $a_n = 5n - 3$, the end terms are $a_{20} = 97$ and $a_{40} = 197$, and the count is $40 - 20 + 1 = 21$.

$S = \frac{21(97 + 197)}{2} = 21 \times 147 = 3087$

The rows run from $25$ down to $10$, a count of $25 - 10 + 1 = 16$ terms, with first term $25$ and last term $10$.

$S = \frac{16(25 + 10)}{2} = 8 \times 35 = 280$

The rows hold $1 + 2 + \cdots + 20$, an arithmetic series with $n = 20$, $a_1 = 1$, and $a_{20} = 20$.

$S_{20} = \frac{20(1 + 20)}{2} = 10 \times 21 = 210$

Use $S_n = \frac{n}{2}(2a_1 + (n-1)d) = \frac{n}{2}(4 + 3(n-1)) = 100$, which becomes $n(3n + 1) = 200$.

Solve $3n^2 + n - 200 = 0$; the positive whole-number root is

$n = \frac{-1 + \sqrt{2401}}{6} = \frac{-1 + 49}{6} = 8$

From $S_n = \frac{n(a_1 + a_n)}{2}$, we get $832 = \frac{n(4 + 100)}{2} = 52n$.

$n = \frac{832}{52} = 16$

Substituting $k = 3, 4, 5, 6, 7$ into $2k + 1$ gives the terms $7, 9, 11, 13, 15$.

$S = \frac{5(7 + 15)}{2} = 5 \times 11 = 55$

With $a_n = 2n + 1$, the end terms are $a_{11} = 23$ and $a_{20} = 41$, and the count is $20 - 11 + 1 = 10$.

$S = \frac{10(23 + 41)}{2} = 5 \times 64 = 320$

Use $S_n = \frac{n}{2}(2a_1 + (n-1)d) = \frac{n}{2}(4 + 3(n-1)) = 155$, which becomes $n(3n + 1) = 310$.

Solve $3n^2 + n - 310 = 0$; the positive whole-number root is

$n = \frac{-1 + \sqrt{3721}}{6} = \frac{-1 + 61}{6} = 10$

The terms run $50, 45, \ldots, 5$, stepping by $5$, so the count is $\frac{50 - 5}{5} + 1 = 10$.

$S = \frac{10(50 + 5)}{2} = 5 \times 55 = 275$