The sum of the first and third of three consecutive integers is 118. What is the value of the second integer?
If $n$ is the second integer, then the first integer is $n-1$ and the third integer is $n+1$.  The sum of the first and third integers is $2n$, so $n=118/2=\boxed{59}$.
59
Level 2
Algebra
Problem: The sum of the first and third of three consecutive integers is 118. What is the value of the second integer?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for the second integer, let the three consecutive integers be represented as x, x+1, and x+2. The first integer is x and the third integer is x+2. According to the problem, the sum of the first and third integers is given as 118. Therefore, we can set up the equation:

x + (x + 2) = 118.

This simplifies to:

2x + 2 = 118.

Next, we isolate the term with x by subtracting 2 from both sides:

2x = 116.

Now, we divide both sides by 2 to solve for x:

x = 58.

Having found the value of the first integer, we can now determine the second integer, which is x + 1:

58 + 1 = 59.

Thus, the value of the second integer is 59.<|END_OF_LATENT|> The final answer is $59$.