Finding Consecutive Integers for Sqrt(290) 

The question asks us to identify the two consecutive integers between which the value of $\sqrt{290}$ lies. To solve this, we need to find an integer $n$ such that $n < \sqrt{290} < n+1$. An equivalent way to think about this is to find an integer $n$ such that $n^2 < 290 < (n+1)^2$.

Estimating the Square Root

We can estimate the value of $\sqrt{290}$ by finding the squares of integers. We are looking for two perfect squares that surround the number 290.

Calculating Integer Squares

Let's calculate the squares of integers to find the range:

Determining the Consecutive Integers

From the calculations above, we observe that:

• $17^2 = 289$
• $18^2 = 324$

Since 290 is greater than 289 (which is $17^2$) and less than 324 (which is $18^2$), we can write the inequality:

$17^2 < 290 < 18^2$

Taking the square root of each part of the inequality, we get:

$\sqrt{17^2} < \sqrt{290} < \sqrt{18^2}$

$17 < \sqrt{290} < 18$

This shows that $\sqrt{290}$ lies between the consecutive integers 17 and 18.

Conclusion

Therefore, $\sqrt{290}$ lies between the consecutive integers 17 and 18. This corresponds to option 3.