What is the sum of all of the integers between the square root of 25 and the square root of 140 ?

Guest Sep 23, 2017

edited by
Guest
Sep 23, 2017

#1**+1 **

\(\sqrt{25}=5\)

Since we only need to calculate the integers, the approximation for \(\sqrt{140}\) does not need to be too accurate. I know that \(12^2=144\) and \(11^2=121\), so \(\sqrt{140}\) can be rounded down to 11.

There is a formula that allows one to calculate the running total of consecutive integers from one to the last integer. It is \(\frac{n(n+1)}{2}\), where *n* is the largest number in the set, in this case 11. However, we have a bit of an issue.

This formula only works when the consecutive integers start at 1. In this case, the integers start at 5. Therefore, we must subtract the sum of the integers from 1 to 4. We can use the same formula of \(\frac{n(n+1)}{2}\), where n=4.

\(\frac{11(11+1)}{2}-\frac{4(4+1)}{2}\)

Find the result of this.

\(\frac{11(11+1)}{2}-\frac{4(4+1)}{2}\) | Do what is inside the parentheses first. |

\(\frac{11*12}{2}-\frac{4*5}{2}\) | |

\(11*6-2*5\) | |

\(66-10\) | |

\(56\) | |

TheXSquaredFactor
Sep 23, 2017