Number and Algebra - Arithmetic sequences

Find the $20^{th}$ term of the arithmetic sequence: $5, 12, 19, 26, \dots$

$u_{20} = 5 + (20 - 1) \times 7$ $u_{20} = 5 + 19 \times 7$ $u_{20} = 5 + 133 = 138$

First, identify the first term $u_1 = 5$. Find the common difference by subtracting the first term from the second: $d = 12 - 5 = 7$. Use the general term formula $u_n = u_1 + (n - 1)d$ with $n = 20$.

Calculate the sum of the first $15$ terms of the sequence: $100, 94, 88, \dots$

$S_{15} = \frac{15}{2}(2(100) + (15 - 1)(-6))$ $S_{15} = 7.5(200 + 14 \times -6)$ $S_{15} = 7.5(200 - 84)$ $S_{15} = 7.5(116) = 870$

Identify $u_1 = 100$ and $d = 94 - 100 = -6$. Since we want the sum of the first $n=15$ terms and we do not know the last term, we use the formula $S_n = \frac{n}{2}(2u_1 + (n - 1)d)$.

In an arithmetic sequence, the $4^{th}$ term is $15$ and the $9^{th}$ term is $35$. Find the first term $u_1$ and the common difference $d$.

$u_4 = u_1 + 3d = 15$ $u_9 = u_1 + 8d = 35$ Subtract the first equation from the second: $(u_1 + 8d) - (u_1 + 3d) = 35 - 15$ $5d = 20 \implies d = 4$ Substitute $d = 4$ back into the first equation: $u_1 + 3(4) = 15$ $u_1 + 12 = 15 \implies u_1 = 3$

Create two simultaneous equations using the $n$th term formula. Subtracting the equations eliminates $u_1$, allowing you to solve for $d$. Then substitute $d$ back to find $u_1$.