sum of arithmetic sequence

**muddywaters** · November 7th 2012, 07:06 AM

I'm sorry if this is in the wrong subforum, I am not sure where to put it. But I searched this topic and saw others posting in pre-calc so im posting this here.

Given the arithmetic sequence $-5,a_1,a_2, ... ,a_n,15$ , and the sum of its first $n+2$ terms is $100$ , find $n$ .

the answer is this:
$S=\frac{(n+2)(-5+15)}{2}=5n+10=100$
$\therefore n=18$

My question is this:
the $\frac{(n+2)(-5+15)}{2}$ part is from $\frac{n(a+l)}{2}$ , where $n$ is the number of terms, $a$ is the first term and $l$ is the last term.

In this case, for the sum of $n+2$ terms, the last term is the $(n+2)_t_h$ term. the answer uses $15$ , which is the $(n+1)_t_h$ term, since it's directly after the $n_t_h$ term. it's not the $(n+2)_t_h$ term.

why is it like that? did i misunderstand something??

**Siron** · November 7th 2012, 07:18 AM

$15$ is the $(n+2)$ th term because $-5$ is the first term, $a_1$ the second, ..., $a_n$ is the $(n+1)$ th term, hence $15$ is the $(n+2)$ th term.

**Plato** · November 7th 2012, 07:24 AM

Originally Posted by muddywaters

Given the arithmetic sequence $-5,a_1,a_2, ... ,a_n,15$ , and the sum of its first $n+2$ terms is $100$ , find $n$ .
My question is this:
the $\frac{(n+2)(-5+15)}{2}$ part is from $\frac{n(a+l)}{2}$ , where $n$ is the number of terms, $a$ is the first term and $l$ is the last term.
In this case, for the sum of $n+2$ terms, the last term is the $(n+2)_t_h$ term. the answer uses $15$ , which is the $(n+1)_t_h$ term, since it's directly after the $n_t_h$ term. it's not the $(n+2)_t_h$ term. why is it like that? did i misunderstand something??

In the sequence $-5,a_1,a_2, ... ,a_n,15$ there are $(n+2)$ terms.
The first term is $-5$ .
The second term is $a_1$ .
$\;~\;~ \vdots$
The last term is $15$ .

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