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Answer :
Method 1:
[tex]F(1)=4\\\\F(n)=F(n-1)+7\\\\n\in\{2,\ 3,\ 4,\ 5,\ ...\}\\\\F(2)=F(1)+7\to F(2)=4+7=11\\\\F(3)=F(2)+7\to F(3)=11+7=18\\\\F(4)=F(3)+7\to F(4)=18+7=25\\\\F(5)=F(4)+7\to F(5)=25+7=32\\\\F(6)=F(5)+7\to F(6)=32+7=39\\\\F(7)=F(6)+7\to F(7)=39+7=46\\\\F(8)=F(7)+7\to F(8)=46+7=53[/tex]
Answer: B. 53
Method 2:
The general formula of an arithmetic sequence:
[tex]a_n=a_1+(n-1)d[/tex]
We have:
[tex]a_1=F(1)=4,\ d=7[/tex]
Substitute:
[tex]a_n=4+(n-1)(7)=4+7n-7=7n-3[/tex]
Put n = 8 to the formula:
[tex]a_8=7(8)-3=56-3=53[/tex]
Answer: B. 53
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