Arithmetic Series: Practice

Here $a_1 = 2$ and $d = 3$. First find the $30$th term: $a_{30} = 2 + 29 \times 3 = 89$.

Now use $S_n = \frac{n(a_1 + a_n)}{2}$ with $n = 30$.

$S_{30} = \frac{30(2 + 89)}{2} = 15 \times 91 = 1365$

The summand $3k - 1$ gives the arithmetic terms $2, 5, 8, 11, 14, 17$ for $k = 1$ to $6$.

Add them with $S_n = \frac{n(a_1 + a_n)}{2}$, using $n = 6$, first term $2$, last term $17$.

$S_6 = \frac{6(2 + 17)}{2} = 3 \times 19 = 57$

The index runs from $5$ to $20$, so use the count $n - m + 1$ with both endpoints included.

$20 - 5 + 1 = 16$

There are $16$ terms.

There are $50$ terms, from $a_1 = 1$ to $a_{50} = 99$.

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

The rows form an arithmetic sequence with $a_1 = 20$ and $d = 3$. The back row is $a_{12} = 20 + 11 \times 3 = 53$.

Sum the $12$ rows with $S_n = \frac{n(a_1 + a_n)}{2}$.

$S_{12} = \frac{12(20 + 53)}{2} = 6 \times 73 = 438$

Substitute $n = 15$, $a_1 = 3$, and $d = 2$ into the second form directly.

$S_{15} = \frac{15}{2}(2 \times 3 + 14 \times 2) = \frac{15}{2}(6 + 28) = 15 \times 17 = 255$

Use $S_n = \frac{n(a_1 + a_n)}{2}$ with $n = 25$, $a_1 = 1$, and $a_{25} = 25$.

$S_{25} = \frac{25(1 + 25)}{2} = \frac{25 \times 26}{2} = 325$

The summand $2k - 1$ gives the odd numbers $1, 3, \ldots, 19$ for $k = 1$ to $10$.

$S_{10} = \frac{10(1 + 19)}{2} = 5 \times 20 = 100$

The rows hold $1 + 2 + \cdots + 15$, an arithmetic series with $n = 15$, $a_1 = 1$, and $a_{15} = 15$.

$S_{15} = \frac{15(1 + 15)}{2} = \frac{15 \times 16}{2} = 120$

The twelfth term is $a_{12} = 4 + 11 \times 4 = 48$.

$S_{12} = \frac{12(4 + 48)}{2} = 6 \times 52 = 312$

There are $10$ multiples of $3$ from $3$ to $30$, so $n = 10$, $a_1 = 3$, and $a_{10} = 30$.

$S_{10} = \frac{10(3 + 30)}{2} = 5 \times 33 = 165$

The fortieth odd number is $a_{40} = 1 + 39 \times 2 = 79$.

$S_{40} = \frac{40(1 + 79)}{2} = 20 \times 80 = 1600$