How do I find the sum of an arithmetic sequence on a calculator?

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How do I find the sum of an arithmetic sequence on a calculator?

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Answer 1 · 2018-03-27T15:52:26

You can use the following general formula:

$n \sum k = 1 (a k + b) = \frac{n}{2} [a (n + 1) + 2 b]$ $sum_(k=1)^n(ak+b)=n/2[a(n+1)+2b]$

where the series has $n$ $n$ terms $a$ $a$ and $b$ $b$ are constants.

(Ask a general question, get a general answer.)

Explanation:

Seems that it would be handy to have a closed form expression for the following:

$n \sum k = 1 (a k + b)$ $sum_(k=1)^n(ak+b)$

where a and b are constants.

The commutative property of addition says

$n \sum k = 1 (a k + b) = n \sum k = 1 a k + n \sum k = 1 b$ $sum_(k=1)^n(ak+b)=sum_(k=1)^nak+sum_(k=1)^nb$

We can factor the $a$ $a$ .

$n \sum k = 1 (a k + b) = a n \sum k = 1 k + n \sum k = 1 b$ $sum_(k=1)^n(ak+b)=asum_(k=1)^nk+sum_(k=1)^nb$

But we know closed form representations for both of these sums.

$n \sum k = 1 k = \frac{n (n + 1)}{2}$ $sum_(k=1)^nk=(n(n+1))/2$

and

$n \sum k = 1 b = n b$ $sum_(k=1)^nb=nb$

So

$n \sum k = 1 (a k + b) = \frac{a n (n + 1)}{2} + n b = \frac{n}{2} [a (n + 1) + 2 b]$ $sum_(k=1)^n(ak+b)=(an(n+1))/2+nb=n/2[a(n+1)+2b]$

EXAMPLE

Suppose we wanted the following sum:

$7 + 10 + 13 + 16 + 19 + 22 + 25 + 28 + 31$ $7+10+13+16+19+22+25+28+31$

There are 9 terms in this series so $n = 9$ $n=9$ . Consecutive terms differ by 3 so $a = 3$ $a=3$ . The first term, $x_{1} = 7$ $x_1=7$ and so $b = x_{1} - a = 7 - 3 = 4$ $b=x_1-a=7-3=4$ .

We want to find

$9 \sum k = 1 (3 k + 4) = \frac{9}{2} [3 (9 + 1) + 2 \cdot 4] = 171$ $sum_(k=1)^9(3k+4)=9/2[3(9+1)+2*4]=171$

which is, in fact, the sum of the numbers given.

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How do I find the sum of an arithmetic sequence on a calculator?

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