You're on a roll — 1. Step 1: Set up equations for the given terms of the arithmetic sequence. The general term of an arithmetic sequence is $T_n = a + (n-1)d$. Given $T_{13} = 15$ and $T_7 = 51$. $$T_{13} = a + (13-1)d \implies a + 12d = 15 \quad (1)$$ $$T_7 = a + (7-1)d \implies a + 6d = 51 \quad (2)$$ Step 2: Solve the system of equations to find $a$ and $d$. Subtract equation (2) from equation (1): $$(a + 12d) - (a + 6d) = 15 - 51$$ $$6d = -36$$ $$d = -6$$ Substitute $d = -6$ into equation (2): $$a + 6(-6) = 51$$ $$a - 36 = 51$$ $$a = 51 + 36$$ $$a = 87$$ The first term is $a=87$ and the common difference is $d=-6$. The general term of the sequence is $T_n = 87 + (n-1)(-6) = 87 - 6n + 6 = 93 - 6n$. 1.a) Step 1: Set $T_n = -21$ and solve for $n$. $$93 - 6n = -21$$ $$-6n = -21 - 93$$ $$-6n = -114$$ $$n = \frac{-114}{-6}$$ $$n = 19$$ The term equal to $-21$ is the $\boxed{\text{19th term}}$. 1.b) Step 1: Set $T_n = 66$ and solve for $n$. $$93 - 6n = 66$$ $$-6n = 66 - 93$$ $$-6n = -27$$ $$n = \frac{-27}{-6}$$ $$n = \frac{9}{2}$$ $$n = 4.5$$ Since $n$ is not an integer, $\boxed{\text{66 is not a term of the sequence}}$. 2. Step 1: Identify the given values for the arithmetic series. First term $a = 5$. Last term $L = 61$. Sum of all terms $S_n = 957$. Step 2: Use the formula for the sum of an arithmetic series. The formula for the sum of an arithmetic series is $S_n = \frac{n}{2}(a + L)$. Substitute the given values into the formula: $$957 = \frac{n}{2}(5 + 61)$$ $$957 = \frac{n}{2}(66)$$ $$957 = 33n$$ Step 3: Solve for $n$. $$n = \frac{957}{33}$$ $$n = 29$$ The number of terms in the series is $\boxed{29}$. Got more? Send 'em!