Find the sum: 11+12+13+….+40

asked by guest
on Dec 08, 2023 at 4:18 am

You asked:

Find the sum of the sequence: \(11\), \(12\), \(13\), \(\ldots\), \(40\)

MathBot Answer:

The sum of the sequence is \(765\)


This is an arithmetic sequence.

The nth term in this sequence is given by the formula:

Explicit Formula: $$a_n=n + 10$$

Recursive Formula: $$a_n=a_{n-1} + 1, \text{where } a_1=11$$


Summation Formula:

Option 1:

$$S_n=\frac{n}{2} (2 a_1 + (n - 1) d)$$ where $a_1$ is the 1st term, $d$ is the common difference, and $n$ is the is the term number.

Option 2:

$$S_n=n \left(\frac{a_1 + a_n}{2}\right)$$ where $a_1$ is the 1st term, $a_n$ is the nth term, and $n$ is the is the term number.

Option 3:

$$\begin{aligned} S_n&=\sum_{i=1}^{n} a_{n} \\ &=\sum_{i=1}^{n} \left(i + 10\right) \\ &=\frac{n \left(n + 21\right)}{2} \end{aligned}$$where $n$ is the is the term number.

Note: Replace all $n$ in $a_n$ in the summation with $i$.


Use the explicit formula to find the index of the final term in the sequence.$$\begin{aligned}40 &= n + 10 \\ n &= 30\end{aligned}$$Then use the summation formula to solve.$$\begin{aligned}S_n &= \frac{n \left(n + 21\right)}{2} \\S_{30} &= \frac{30 \cdot \left(21 + 30\right)}{2}\\S_{30} &= 765\end{aligned}$$

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