The sum of the first n terms of a sequence is n^2 + 5n. What is the 1000th term of the sequence?

Guest May 7, 2020

#1**0 **

The list of numbers is just even numbers starting at 6, so the 1000th term would be **2006**.

HELPMEEEEEEEEEEEEE May 7, 2020

#2**0 **

The sequence is:

3, 6, 9, 12, 15, 18, 21 =n^2 + 5*n=7^2 + 35=84

Therefore the 1000th term =First term + 3*(1000 - 1)

=3 + (3*999)

=3 + 2,997

**=3,000 - which is 1000th term.**

Guest May 7, 2020

edited by
Guest
May 7, 2020

#3**0 **

The sum of the first term is just the first term. So if we plug in 1 to the equation, we'll find the first term which is (1^2) + 5 * 1 = 6. From this I can already see the difference in our solutions.

HELPMEEEEEEEEEEEEE
May 7, 2020

#5**+1 **

**The sum of the first n terms of a sequence is \(n^2 + 5n\). What is the \(1000th\) term of the sequence?**

**Formula: **\(\mathbf{\boxed{s_n =\dfrac{(a_1+a_n)}{2}\cdot n }}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_1} &=& \mathbf{s_1} \quad | \quad s_n =n^2 + 5n \\ &=& 1^2 + 5*1 \\ \mathbf{a_1} &=& \mathbf{6} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_n &=& \dfrac{(a_1+a_n)}{2}\cdot n \quad | \quad a_1=6,\ s_n = n^2 + 5n \\\\ n^2 + 5n &=& \dfrac{(6+a_n)}{2}\cdot n \\\\ 2n^2 + 10n &=& (6+a_n)n \quad | \quad :n \\ 2n + 10 &=& 6+a_n \\ 2n + 4 &=& a_n \\ \mathbf{a_n} &=& \mathbf{2n+4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_n} &=& \mathbf{2n+4} \quad | \quad n=1000 \\ a_{1000} &=& 2*1000+4 \\ \mathbf{a_{1000}}&=& \mathbf{2004} \\ \hline \end{array}\)

heureka May 8, 2020