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Let be $a_1$, $a_2$, $\ldots$, $a_{n-1}$, $a_n$, term of the arithmetics sequence. Please help me prove that

$S_n=\frac{n}{2}(a_1+a_n)$

is sum the n-th term of the given sequence. Thanks.

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HINT: $a_1+a_n=a_2+a_{n-1}=\ldots$ –  Raskolnikov Oct 10 '12 at 7:57
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2 Answers

up vote 0 down vote accepted

Let be $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ real numerical sequence. That one sequence let be arithmetics sequence must that

$a_2-a_1=a_3-a_2=\cdots=d$.

Note that the given sequence is

$2-1=3-2=4-3=5-4=\cdots=1$,

therefore the given sequence is arithmetics sequence. Also note that

$1+11=2+9=3+8=\cdots$.

So generally worth

$(*)$ $\ldots$ $a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\cdots$

Now turn prove the given formula.

The given sequence is

$a_1, a_2, \ldots, a_{n-1}, a_n$.

Form the sum of the given sequence, and that sum the mark $S_n$, i.e

$(1)$ $\ldots$ $S_n=a_1+a_2+\ldots+a_{n-1}+a_n$

Because comutative low for the sum (+) is worth, have

$(2)$ $\ldots$ $S_n=a_n+a_{n-1}+\ldots+a_2+a_1$

Equation $(1)$, $(2)$ collect side to side

$S_n+S_n=a_1+a_2+\ldots+a_{n-1}+a_n+a_n+a_{n-1}+\ldots+a_2+a_1$

$2S_n=(a_1+a_n)+(a_2+a_{n-1})+\cdots+(a_{n-1}+a_2)+(a_n+a_1)$

Implement the $(*)$ we have:

$2S_n=(a_1+a_n)+(a_1+a_n)+\ldots+(a_1+a_n)$

$2S_n=n(a_1+a_n)$$/$$:2$

$S_n=\frac{n}{2}(a_1+a_n)$

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Thanky sir, for proving –  user39471 Oct 10 '12 at 8:11
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How did you manage to type this up in less than 3 minutes? –  commenter Oct 10 '12 at 8:21
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Writing the arithmetic sequence in another way, $$a_0 + (a_0 + d) + (a_0 + 2d) + ...$$

Flip it around to get another sequence that decreases to $a_0$,

$$ (a_0 + nd) + (a_0 + (n-1)d) + ...$$

Notice that the sum of the $k$-th term of one sequence with the other is always $(2a_0 + nd) = (a_0 + a_n)$.

There are $n$ such terms. So $n(a_0 + a_n)$ is twice the sum of the sequence.

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