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elcharly64

Answer:

[tex]S_{21}=8.4[/tex]

Step-by-step explanation:

Sum Of Arithmetic Sequences

Given a sequence

[tex]a_1, a_1+r, a_1+2r, ...., a_1+(n-1)r[/tex]

The  sum of all terms is

[tex]\displaystyle S_n=n\frac{a_1+a_n}{2}[/tex]

[tex]a_n=a_1+(n-1)r[/tex]

If we know [tex]a_n, a_1, r[/tex]

we can compute n as

[tex]\displaystyle n=\frac{a_n-a_1}{r}+1[/tex]

The given sequence is

[tex]0.30+0.31+0.32+...+0.50[/tex]

The common difference is

[tex]r=0.31-0.30=0.01[/tex]

[tex]a_1=0.30, a_n=0.50[/tex]

We compute n

[tex]\displaystyle n=\frac{0.50-0.30}{0.01}+1[/tex]

n=21

So the given sum is

[tex]\displaystyle S_{21}=21\frac{0.30+0.50}{2}[/tex]

[tex]S_{21}=21(0.40)[/tex]

[tex]S_{21}=8.4[/tex]

Athenalsun

Answer:

8.4

Step-by-step explanation:

We've seen problems like this with integers. Even though there are decimals here, the strategy is the same. We pair off terms of the sum from either end:

0.30+0.50, 0.31+0.49,  0.32+0.48, 0.33+0.47, and so on.

If we continue like this, then each pair has the same sum, 0.80. However, the middle term, 0.40, will be left over with no partner. There are 10 pairs, so the total of all 21 numbers is (10) x (0.8) + 0.4. This is equal to 8.4

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