Sequences and Series

$a = 3$ and $d = 4$, so $a_{20} = 3 + 19(4) = 3 + 76 = 79$. There are 19 gaps between 20 terms.

$a=2$, $d=3$, $n=15$. $S_{15} = \tfrac{15}{2}\big[4 + 14(3)\big] = \tfrac{15}{2}(46) = 345$.

$a = 5$ and $r = 2$, so $a_7 = 5 \times 2^6 = 5 \times 64 = 320$.

$\dfrac{a}{1 - \tfrac14} = 40$, so $\dfrac{a}{\tfrac34} = 40$ and $a = 40 \times \tfrac34 = 30$.

$a + 2d = 11$ and $a + 7d = 26$, so $5d = 15$, giving $d = 3$ and $a = 5$. Then $S_{20} = \tfrac{20}{2}\big[10 + 19(3)\big] = 10(67) = 670$.

Writing out the first terms gives $1, 4, 7, \ldots$, an arithmetic progression with $a=1$, $d=3$ and $n=12$. So $S_{12} = \tfrac{12}{2}\big[2 + 11(3)\big] = 6(35) = 210$.