Respuesta :
Using the sum of the first n terms of an arithmetic sequence, it is found that the possible values of k are given by: k = 0.73, k = 30.27.
What is an arithmetic sequence?
In an arithmetic sequence, the difference between consecutive terms is always the same, called common difference d.
The nth term of an arithmetic sequence is given by:
[tex]a_n = a_1 + (n - 1)d[/tex]
In which [tex]a_1[/tex] is the first term.
The sum of the first n terms is given by:
[tex]S_n = \frac{n(a_1 + a_n)}{2}[/tex]
In this problem, the sequence is given by:
[tex]a_n = 48 - 3n[/tex]
Hence:
- [tex]a_1 = 45, d = -3[/tex]
- [tex]a_k = 48 - 3k[/tex]
We want the sum to be equals to 330, hence:
[tex]S_n = \frac{n(a_1 + a_n)}{2}[/tex]
[tex]330 = \frac{k(45 + 48 - 3k)}{2}[/tex]
[tex]-3k^2 + 93k = 660[/tex]
[tex]3k^2 - 93k + 660 = 0[/tex]
[tex]k^2 - 31k + 22 = 0[/tex]
Which is a quadratic function with coefficients a = 1, b = -31, c = 22, hence:
[tex]\Delta = b^2 - 4ac = (-31)^2 - 4(1)(22) = 873[/tex]
[tex]x_1 = \frac{31 + \sqrt{873}}{2} = 30.27[/tex]
[tex]x_2 = \frac{31 - \sqrt{873}}{2} = 0.73[/tex]
More can be learned about arithmetic sequences at https://brainly.com/question/6561461
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Answer:
k = 11 or 20
Step-by-step explanation:
The sum of terms of an arithmetic sequence is given by ...
Sn = (a1 +an)(n/2)
We want to find the value of n for a particular series to have a particular sum.
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parameters
The series of interest is defined by ...
an = 48 -3n
so, the first term is ...
a1 = 48 -3(1) = 45
The k-th term is ...
ak = 48 -3k
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sum
Then the sum is ...
Sk = (45 +(48 -3k))(k/2)
We want this value to be 330, so we have ...
330 = (93 -3k)(k/2)
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solution
220 = (31 -k)k . . . . . . . . multiply by 2/3
k² -31k +220 = 0 . . . . . in standard form
We want factors of 220 that total 31.
220 = (1)(220) = (2)(110) = (4)(55) = (5)(44) = (10)(22) = (11)(20)
The last of these pairs totals 31, so we have ...
(k -11)(k -20) = 0 . . . . . factored form of the equation
k = 11 or k = 20 . . . . . values of k that make the factors zero
Possible values of k are 11 and 20.