Number and Algebra - Arithmetic sequences

Find the $25^{th}$ term of the arithmetic sequence $12, 19, 26, 33, \dots$

$u_1 = 12$ $d = 19 - 12 = 7$ $n = 25$ Using $u_n = u_1 + (n-1)d$: $u_{25} = 12 + (25-1) \times 7$ $u_{25} = 12 + 24 \times 7$ $u_{25} = 12 + 168 = 180$

Identify the first term $u_1$ and calculate the common difference $d$. Substitute these values into the general term formula for $n = 25$.

Calculate the sum of the first $40$ terms of an arithmetic sequence where the first term is $5$ and the common difference is $3$.

$u_1 = 5$ $d = 3$ $n = 40$ Using $S_n = \frac{n}{2}(2u_1 + (n-1)d)$ $S_{40} = \frac{40}{2}(2(5) + (40-1)(3))$ $S_{40} = 20(10 + 39 \times 3)$ $S_{40} = 20(10 + 117)$ $S_{40} = 20(127) = 2540$

Use the arithmetic series formula. Substitute the known values $u_1$, $d$, and $n$ to find the total sum.

In an arithmetic sequence, the $4^{th}$ term is $15$ and the $10^{th}$ term is $39$. Find the first term $u_1$ and the common difference $d$.

We have two equations based on $u_n = u_1 + (n-1)d$:

1. $u_4 = u_1 + 3d = 15$
2. $u_{10} = u_1 + 9d = 39$ Subtract equation (1) from (2): $(u_1 + 9d) - (u_1 + 3d) = 39 - 15$ $6d = 24 \implies d = 4$ Substitute $d = 4$ back into (1): $u_1 + 3(4) = 15$ $u_1 + 12 = 15 \implies u_1 = 3$

Create a system of linear equations using the general term formula for the given terms, then solve for $u_1$ and $d$ using elimination or substitution.