Respuesta :
Answer:
Student can obtain 15 marks by attempting 8 questions.
Step-by-step explanation:
This question is incomplete; complete question is here.
A student taking a test consisting of 10 questions is told that each questions after the first is worth 2 marks more than the preceding question. If the third question of the test is worth 5 marks.What is the maximum score that the student can obtain by attempting 8 questions?
A students when attempts the questions the sequence formed by the scores he gets will be in the form of a, (a + 2), (a + 4), (a + 6)........
Which will be an arithmetic sequence with a common difference = 2
We know explicit formula of an arithmetic sequence is
[tex]T_{n}=a+(n-1)d[/tex]
Where [tex]T_{n}[/tex] = First term of the sequence
a = first term
n = number of term
d = common difference
It is given that 3rd term of this sequence is 5.
[tex]5=a+(3-1)2[/tex]
a = 5 - 4 = 1
Explicit formula for the sequence will be [tex]T_{n}=1+(n-1)2[/tex]
[tex]T_{n}=2n-1[/tex]
Now if the student attempts 8 questions then from the explicit formula
[tex]T_{8}=8\times 2-1[/tex]
[tex]T_{8}=15[/tex]
Therefore, student can obtain 15 marks by attempting 8 questions.
The maximum score that the student can obtain is 19 marks.
Given that each questions after the first is worth 2 marks more than the preceding question.
Let us consider that marks of first question is x , marks of second question is (x + 2) so on.
A series of marks is formed,
[tex]x, (x+2),(x+4),(x+6),........[/tex]
Common difference,[tex]d=2[/tex]
If the third question of the test is worth 5 marks.
So that, [tex]x+4=5\\\\x=5-4=1[/tex]
The series is, [tex]1,3,5,7,.....[/tex]
So that, maximum marks for 10th question,
[tex]a_{n}=a_{1}+(n-1)d\\\\a_{10}=a_{1}+9d=1+9(2)=19[/tex]
the maximum score that the student can obtain is 19 marks.
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