To find the common difference $\displaystyle d$, use the formula $\displaystyle a_{1}+4d=a_{5}$.

For us, $\displaystyle a_1$ is $\displaystyle 10$ and $\displaystyle a_5$ is $\displaystyle 38$.

$\displaystyle 10+4d=38$

Now we can solve for $\displaystyle d$.

$\displaystyle 4d=28$

$\displaystyle d =7$

Add the common difference to the first term to get the second term.

$\displaystyle a_{2}=a_{1}+d=10+7 = 17$

Let $\displaystyle d$ be the common difference, and let $\displaystyle x$ be the second term. The first three terms are, in order, $\displaystyle x-d, x, x+d$.

The sum of the first three terms is $\displaystyle (x-d) + x+(x+d)=111$.

$\displaystyle x+x+x+d-d=3x=111$

$\displaystyle x=\frac{111}{3}=37$

Now we know that the second term is 37. The fourth term is the second term plus twice the common difference: $\displaystyle x+2d$. Since the second and fourth terms are 37 and 49, respectively, we can solve for the common difference.

$\displaystyle x+2d=37+2d=49$

$\displaystyle 2d=12$

$\displaystyle d=6$

The common difference is 6. The first term is $\displaystyle x-d = 37-6=31$.

For arithmetic sequences, we use the formula $\displaystyle a_{n} = a_{1}+(n-1)d$, where $\displaystyle a_{n}$ is the term we are trying to find, $\displaystyle a_{1}$ is the first term, and $\displaystyle d$ is the difference between consecutive terms. In this case, $\displaystyle a_{1} = -9$ and $\displaystyle d = 5$. So, we can write the formula as $\displaystyle a_{31} = -9 +(30)(5)$, and $\displaystyle a_{31} = 141$.

Let $\displaystyle d$ be the common difference of the sequence. Then $\displaystyle a_{4} + 6d = a_{10}$, or, equivalently,

$\displaystyle d = \frac{a_{10} - a_{4}}{6}= \frac{888 - 372}{6} = \frac{516}{6}= 86$

$\displaystyle a_{1} + 3d = a_{4 }$ , or equivalently,

$\displaystyle a_{1} = a_{4 }-3d = 372-3\cdot 86 = 114$

The common difference of the sequence is the difference of the tenth and ninth terms: $\displaystyle d = 99-87 = 12$.

The ninth term of an arithmetic sequence with first term $\displaystyle a$ and common difference $\displaystyle d$ is $\displaystyle a + 8d$, so we set this equal to 87, set $\displaystyle d=12$, and solve:

$\displaystyle a + 8d = 87$

$\displaystyle a + 8 \cdot12 = 87$

$\displaystyle a + 96 = 87$

$\displaystyle a = 87 -96 = -9$

The eighth and tenth terms of the sequence are $\displaystyle a +7d$ and $\displaystyle a + 9d$, where $\displaystyle a$ is the first term and $\displaystyle d$ is the common difference. We can find the common difference by subtracting the tenth and eighth terms and solving for $\displaystyle d$:

$\displaystyle a + 9d = 99$

$\displaystyle \underline{a+7d = 87}$

        $\displaystyle 2d= 12$

$\displaystyle 2d \div 2= 12 \div 2$

$\displaystyle d=6$

Now set eighth term $\displaystyle a + 7d$ equal to 87, set $\displaystyle d=6$, and solve:

$\displaystyle a + 7d = 87$

$\displaystyle a + 7\cdot 6= 87$

$\displaystyle a + 42= 87$

$\displaystyle a + 42 -42= 87 -42$

$\displaystyle a=45$

Start by finding the common difference, $\displaystyle d$, in this sequence, which you can get by subtracting the first term from the second.

$\displaystyle d=5-2=3$

Then, using the formula given before the question:

$\displaystyle a_{18}=2+(18-1)3=2+(17)3=2+51=53$

Start by finding the common difference in terms by subtracting the first term from the second.

$\displaystyle d=92-101=-9$

Then, fill in the rest of the equation given before the question.

$\displaystyle a_{26}=101+(26-1)-9=101+(25)-9$

$\displaystyle =101+-225=-124$

The 11th term means there are 10 gaps in between the first term and the 11th term. Each gap has a difference of +4, so the 11th term would be given by 10 * 4 + 1 = 41.

The first term is 1.

Each term after increases by +4.

The n^th term will be equal to 1 + (n – 1)(4).

The 11th term will be 1 + (11 – 1)(4)

1 + (10)(4) = 1 + (40) = 41