SOLUTION: Algebra 2 worksheet on Arithmetic sequence The third term of an arithmetic sequence is 9 and the seventh term is 31. Find the sum of the first twenty-two terms. How many t

Algebra ->  Sequences-and-series -> SOLUTION: Algebra 2 worksheet on Arithmetic sequence The third term of an arithmetic sequence is 9 and the seventh term is 31. Find the sum of the first twenty-two terms. How many t      Log On


   



Question 202838: Algebra 2 worksheet on Arithmetic sequence
The third term of an arithmetic sequence is 9 and the seventh term is 31. Find the sum of the first twenty-two terms.

How many terms of the arithmetic sequence 2,4,6,8, ... add up to 60,762?

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The third term of an arithmetic sequence is 9 and the seventh term is 31. Find the sum of the first twenty-two terms.

___, ___,  9 , ___, ___, ___,  31, ...

The nth term formula is

a%5Bn%5D=+a%5B1%5D+%2B+%28n-1%29d

Substitute n=3 in

a%5Bn%5D=+a%5B1%5D+%2B+%28n-1%29d

a%5B3%5D=+a%5B1%5D+%2B+%283-1%29d

a%5B3%5D=+a%5B1%5D+%2B+2d

Substitute a%5B3%5D=9

9=+a%5B1%5D+%2B+2d

Now Substitute n=7 in

a%5Bn%5D=+a%5B1%5D+%2B+%28n-1%29d

a%5B7%5D=+a%5B1%5D+%2B+%287-1%29d

a%5B7%5D=+a%5B1%5D+%2B+6d

Substitute a%5B7%5D=31

31=+a%5B1%5D+%2B+6d

So we have the system:

system%289=+a%5B1%5D+%2B+2d%2C+31=+a%5B1%5D+%2B+6d%29

Solve the first equation for a%5B1%5D

9=+a%5B1%5D+%2B+2d
9-2d=a%5B1%5D

Substitute a%5B1%5D=9-2d in the second equation:

31=+a%5B1%5D+%2B+6d
31=+%289-2d%29+%2B+6d
31=+9-2d+%2B+6d
31=+9+%2B+4d
31-9=+4d
22=4d
22%2F4=d
11%2F2=d

Substituting d=11%2F2 in

31=+a%5B1%5D+%2B+6%2811%2F2%29
31=+a%5B1%5D+%2B+33
31-33=a%5B1%5D
-2=a%5B1%5D

Now we can use the formula for the sum of the
first n terms:

S%5Bn%5D=%28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%29

Substitute n=22, a%5B1%5D=-2, d=11%2F2

S%5B22%5D=%2822%2F2%29%282%28-2%29%2B%2822-1%29%2811%2F2%29%29
S%5B22%5D=%2811%29%28-4%2B%2821%29%2811%2F2%29%29
S%5B22%5D=%2811%29%28-4%2B231%2F2%29
S%5B22%5D=%2811%29%28-8%2F2%2B231%2F2%29
S%5B22%5D=%2811%29%28223%2F2%29
S%5B22%5D=2453%2F2

----------------

How many terms of the arithmetic sequence 2,4,6,8, ... add up to 60,762

Subtract any two consecutive terms and get d=2

a%5B1%5D = the first term, which is 2, and S%5Bn%5D=60762

S%5Bn%5D=%28n%2F2%29%282a%5B1%5D%2B%28n-1%29d%29

60762=%28n%2F2%29%282%282%29%2B%28n-1%29%282%29%29

60762=%28n%2F2%29%284%2B2%28n-1%29%29

60762=%28n%2F2%29%284%2B2n-2%29

60762=%28n%2F2%29%282%2B2n%29

Multiply both sides by 2:

2%2A60762=2%2A%28n%2F2%29%282%2B2n%29

121524=cross%282%29%2A%28n%2Fcross%282%29%29%282%2B2n%29

121524=n%282%2B2n%29

121524=2n%2B2n%5E2

2n%5E2%2B2n-121524=0

Divide through by 2

n%5E2%2Bn-60762=0

Factor:

%28n-246%29%28n%2B247%29=0

n-246=0 or n=246
n%2B247=0 or n=-247

We discard the negative answer, and so

n=246

Edwin