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solong [7]
3 years ago
13

An arithmetic series begins 29+25+21+... find the maximum value that the series could have

Mathematics
1 answer:
Fittoniya [83]3 years ago
5 0

The series' value is maximized if the sum only consists of positive terms. Notice that each term in the sum takes the form 4<em>n</em> + 1 for integer <em>n</em>. This means the smallest positive integer that the sum can involve is 1, so the maximum value is

<em>S</em> = 29 + 25 + 21 + … + 9 + 5 + 1

Reversing the order of terms gives the same sum,

<em>S</em> = 1 + 5 + 9 + … + 21 + 25 + 29

Adding terms in the same positions gives us twice this sum,

2<em>S</em> = (29 + 1) + (25 + 5) + (21 + 9) + … + (1 + 29)

Notice how each grouped sum adds to 30. There are 8 terms in the sum, since 4<em>n</em> + 1 = 29 when <em>n</em> = 8. So

2<em>S</em> = 8 × 30 = 240   ===>   <em>S</em> = 120

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Solve forx: 3 A. 6 B. 6&gt;x&gt;9<br> C. 0 D. 0&gt;x&gt;3<br><br> Real answers only Please
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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

|-6|

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Read 2 more answers
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