Find the 48th term of 9,13,17 . Round to the nearest thousandth if necessary.

jerrylinxdd
Dec 19, 2017

#1**+2 **

This one may not be nearly as obvious as the previous problem. The given sequence is definitely an arithmetic sequence because there is a common difference. The general formula for an arithmetic sequence is the following:

\(a_n=d(n-1)+c\)

d = the common difference

c = the first term in the sequence (sometimes referred to as a_{1})

The common difference is the constant amount of change between numbers. In this case, there exists a common differnece.

\(\underbrace{9, 13}\underbrace{, 17},...\\ +4\hspace{3mm}+4\)

Since the common diffence is 4, let's put that in for d.

\(a_n=4(n-1)+c\)

The first term in this sequence is given; it is 9.

\(a_n=4(n-1)+9\)

Just like that, we have generated the formula that generates the nth term in the sequence. To figure out the 48th term, simply evaluate at n=48.

\(a_{48}=4(48-1)+9\\ a_{48}=4*47+9\\ a_{48}=197\)

TheXSquaredFactor
Dec 19, 2017

#1**+2 **

Best Answer

This one may not be nearly as obvious as the previous problem. The given sequence is definitely an arithmetic sequence because there is a common difference. The general formula for an arithmetic sequence is the following:

\(a_n=d(n-1)+c\)

d = the common difference

c = the first term in the sequence (sometimes referred to as a_{1})

The common difference is the constant amount of change between numbers. In this case, there exists a common differnece.

\(\underbrace{9, 13}\underbrace{, 17},...\\ +4\hspace{3mm}+4\)

Since the common diffence is 4, let's put that in for d.

\(a_n=4(n-1)+c\)

The first term in this sequence is given; it is 9.

\(a_n=4(n-1)+9\)

Just like that, we have generated the formula that generates the nth term in the sequence. To figure out the 48th term, simply evaluate at n=48.

\(a_{48}=4(48-1)+9\\ a_{48}=4*47+9\\ a_{48}=197\)

TheXSquaredFactor
Dec 19, 2017