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lys-0071 [83]
3 years ago
5

Calculate S58 for the arithmetic sequence {an}= {5/6n+1/3} If you could show me the steps that would be great. I keep following

the directions in the lesson and getting 8,613/6 but all I have to choose from for answers are
146/3
91/2
8671/6
9267/6
Mathematics
1 answer:
Amanda [17]3 years ago
5 0

Answer is 8671/6 which is the third choice

===================================

Work Shown:

Find the first term of the sequence by plugging in n = 1

a_n = (5/6)*n + 1/3

a_1 = (5/6)*1 + 1/3    replace n with 1

a_1 = 5/6 + 1/3

a_1 = 5/6 + 2/6

a_1 = 7/6

Repeat for n = 58 to get the 58th term

a_n = (5/6)*n + 1/3

a_58 = (5/6)*58 + 1/3    replace n with 58

a_58 = (5/6)*(58/1) + 1/3

a_58 = (5*58)/(6*1) + 1/3

a_58 = 290/6 + 1/3

a_58 = 145/3 + 1/3

a_58 = 146/3

Now we can use the s_n formula below with n = 58

s_n = (n/2)*(a_1 + a_n)

s_58 = (58/2)*(a_1 + a_58)    replace n with 58

s_58 = (58/2)*(7/6 + a_58)   replace a_1 with 7/6

s_58 = (58/2)*(7/6 + 146/3)   replace a_58 with 146/3

s_58 = (58/2)*(7/6 + 292/6)

s_58 = (58/2)*(299/6)

s_58 = (58*299)/(2*6)

s_58 = 17342/12

s_58 = 8671/6


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Answer:

The number of different lab groups possible is 84.

Step-by-step explanation:

<u>Given</u>:

A class consists of 5 engineers and 4 non-engineers.

A lab groups of 3 are to be formed of these 9 students.

The problem can be solved using combinations.

Combinations is the number of ways to select <em>k</em> items from a group of <em>n</em> items without replacement. The order of the arrangement does not matter in combinations.

The combination of <em>k</em> items from <em>n</em> items is: {n\choose k}=\frac{n!}{k!(n-k)!}

Compute the number of different lab groups possible as follows:

The number of ways of selecting 3 students from 9 is = {n\choose k}={9\choose 3}

                                                                                         =\frac{9!}{3!(9 - 3)!}\\=\frac{9!}{3!\times 6!}\\=\frac{362880}{6\times720}\\ =84

Thus, the number of different lab groups possible is 84.

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Need help on this question
Aloiza [94]
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