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Sagot :
Answer:
Step-by-step explanation:
Let say 2*a+1 the first odd integer.
[tex]n_1=2a+1\\n_2=2a+1+2=2a+3=2a+2*1+1\\n_3=2a+5=2a+2*2+1\\n_4=2a+2*3+1\\....\\n_{384}=2a+2*383+1=2a+767\\\\Sum=(2a+1)+(2a+3)+(2a+5)+...+(2a+767)\\=384*2a+(1+3+...+767)\\=384*2a+\dfrac{1+767}{2} 384\\=384*2a+384* 384\\=384*(2a+384)\\\\The\ sum\ must\ be \ divisible \ by\ 384.\\In\ the\ given\ numbers,\ only\ 20736\ is\ divisible\ by\ 384.\\\\20736=54*384\\\\2a+384=54\\2a=54-384\\2a=-330\\a=-165\\[/tex]
[tex]The\ first\ number\ is\ 2*a+1=2*(-165)+1=-329\\The\ last\ number\ is\ 2a+767=2*(-165)+767= 437\\-329-327-325-....+1+3+5+...+437=20736\\[/tex]
Proof in the file jointed.
Using an arithmetic sequence, the smallest possible sum is of 20 736, given by option b.
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The nth term of an arithmetic sequence is given by:
[tex]a_n = a_1 + (n-1)d[/tex]
In which
- The first term is [tex]a_1[/tex]
- The common ratio is [tex]d[/tex].
The sum of the first n terms of a sequence is given by:
[tex]S_n = \frac{n(a_1 + a_n)}{2}[/tex]
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- 384 means that [tex]n = 384[/tex], thus:
[tex]S_n = \frac{384(a_1 + a_n)}{2}[/tex]
[tex]S_n = 192(a_1 + a_n)[/tex]
- Consecutive odd integers means that the common ratio is 2, thus [tex]d = 2[/tex], and the nth term is:
[tex]a_n = a_1 + (n-1)d = a_1 + 383(2) = a_1 + 766[/tex]
Thus, the sum is:
[tex]S_n = 192(a_1 + a_n)[/tex]
[tex]S_n = 192(a_1 + a_1 + 766)[/tex]
[tex]S_n = 384a_1 + 147072[/tex]
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- Now, for each sum, it is tested if [tex]a_1[/tex] is an odd integer.
For a sum of 104 976.
[tex]S_n = 384a_1 + 147072[/tex]
[tex]104976 = 384a_1 + 147072[/tex]
[tex]384a_1 = 104976 - 147072[/tex]
[tex]a_1 = \frac{104976 - 147072}{384}[/tex]
[tex]a_1 = -109.6[/tex]
Not an integer, so this option is not correct.
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For a sum of 20 736.
[tex]S_n = 384a_1 + 147072[/tex]
[tex]20736 = 384a_1 + 147072[/tex]
[tex]384a_1 = 20736 - 147072[/tex]
[tex]a_1 = \frac{20736 - 147072}{384}[/tex]
[tex]a_1 = -329[/tex]
Odd integer, thus, the correct option is b.
A similar problem is given at https://brainly.com/question/24555380
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