Arithmetic Series: Practice

Use $S_n = \frac{n}{2}(2a_1 + (n-1)d)$, so $\frac{n}{2}(6 + 5(n-1)) = 255$, which becomes $n(5n + 1) = 510$.

Solve $5n^2 + n - 510 = 0$; the positive whole-number root is

$n = \frac{-1 + \sqrt{10201}}{10} = \frac{-1 + 101}{10} = 10$

Use $1 + 2 + \cdots + n = \frac{n(n+1)}{2} = 210$, so $n(n+1) = 420$.

Look for consecutive factors of $420$, which are $20$ and $21$.

$n(n+1) = 20 \times 21 = 420 \;\Rightarrow\; n = 20$

The index runs from $10$ to $20$, so the count is $20 - 10 + 1 = 11$, with first term $10$ and last term $20$.

$S = \frac{11(10 + 20)}{2} = 11 \times 15 = 165$

From $S_n = \frac{n(a_1 + a_n)}{2}$, we get $380 = \frac{n(8 + 68)}{2} = 38n$.

Divide both sides by $38$.

$n = \frac{380}{38} = 10$

The odd numbers run from $a_1 = 1$ to $a_n = 2n - 1$, so $S_n = \frac{n(1 + (2n-1))}{2} = \frac{n \times 2n}{2}$.

$S_n = \frac{2n^2}{2} = n^2$

Both end terms and the count are given, so use $S_n = \frac{n(a_1 + a_n)}{2}$ with $n = 20$.

$S_{20} = \frac{20(15 + 53)}{2} = 10 \times 68 = 680$

The even numbers step by $2$, so the count is $\frac{60 - 20}{2} + 1 = 21$, with first term $20$ and last term $60$.

$S = \frac{21(20 + 60)}{2} = 21 \times 40 = 840$

The twentieth term is $a_{20} = 5 + 19 \times 4 = 81$.

$S_{20} = \frac{20(5 + 81)}{2} = 10 \times 86 = 860$

With $a_n = 2n - 1$, the end terms are $a_6 = 11$ and $a_{15} = 29$, and the count is $15 - 6 + 1 = 10$.

$S = \frac{10(11 + 29)}{2} = 5 \times 40 = 200$

From $\frac{n(n+1)}{2} = 136$ we get $n(n+1) = 272$. Look for consecutive factors of $272$.

$n(n+1) = 16 \times 17 = 272 \;\Rightarrow\; n = 16$

The seats form an arithmetic sequence with $a_1 = 12$ and $d = 3$, so the last row is $a_{25} = 12 + 24 \times 3 = 84$.

$S_{25} = \frac{25(12 + 84)}{2} = 25 \times 48 = 1200$

Substitute to get $0 = \frac{9}{2}(2(-8) + 8d) = \frac{9}{2}(-16 + 8d)$, so $-16 + 8d = 0$.

$8d = 16 \;\Rightarrow\; d = 2$