SOLUTION: I have to use one of the arithmetic series formula. The arithmetic series 5 + 9 + 13 +...+ tn has a sum of 945. How many terms does this series have?

Algebra ->  Sequences-and-series -> SOLUTION: I have to use one of the arithmetic series formula. The arithmetic series 5 + 9 + 13 +...+ tn has a sum of 945. How many terms does this series have?      Log On


   

Question 995635: I have to use one of the arithmetic series formula. The arithmetic series 5 + 9 + 13 +...+ tn has a sum of 945. How many terms does this series have?
Found 2 solutions by MathLover1, atique.ah:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

5, 9, 13,+...+ t%5Bn%5D has a sum of 945


I can see from the sequence 5, 9, 13, ... that each term is 4+more than the previous term.
The first term is 5.
The second term is 9=5%2B4.
The third term is 2 fours plus 5: 13=5%2B2%2A4.
Thus the nth term must be n-1 fours plus 5.
That is +t%5Bn%5D+=+4%28n+-+1%29+%2B+5+=+1+%2B+4n.
Thus the n term series is:
S+=+5+%2B+9+%2B+13 + ... ++%281+%2B+4%28n-1%29%29+%2B+%281+%2B+4n%29
You can find the sum of this series by writing it forward and backwards and then adding down
S=+5+%2B+9++ ... %2B+1%2B4%28n-1%29+%2B+1%2B4n
S+=+1%2B4n+%2B+1%2B4%28n-1%29++ ... + 9 + 5}}}
2S+=+6%2B4n+%2B+6%2B4n++ ... + 6%2B4n+%2B+6%2B4n
Since there are n terms in the series, we have
2S+=+n%286+%2B+4n%29
and thus
S+=+n%286+%2B+4n%29%2F2+
since given the sum
+n%286+%2B+4n%29%2F2+=+945 solve for n to find how many terms are in this sequence
+n%286+%2B+4n%29+=+945%2A2
+6n+%2B+4n%5E2=+1890
++4n%5E2%2B6n+-1890=0
n+=+-45%2F2....disregard negative solution
n+=+21
so, there is 21 terms in your sequence, and they are
5, 9, 13,17, 21, 25,29, 33, 37,41, 45, 49,53, 57, 61,65, 69, 73,77, 81, 85
check their sum:

945=945


Answer by atique.ah(27) About Me  (Show Source):
You can put this solution on YOUR website!
Sn=(t1+tn)*(n/2)
945=(5+tn)*(n/2)
945*2=(5+tn)*n
1890=(5+tn)*n .......... 1
tn=t1+(n-1)d
tn=5+(n-1)4
tn=5+4n-4
tn=1+4n ................2
substituting the value of tn in eq 1
1890=(5+1+4n)*n
1890=(6+4n)*n
1890=6n+4n^2 or writing in the standard form
4n^2+6n-1890=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation an%5E2%2Bbn%2Bc=0 (in our case 2n%5E2%2B3n%2B-945+=+0) has the following solutons:

n%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%283%29%5E2-4%2A2%2A-945=7569.

Discriminant d=7569 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-3%2B-sqrt%28+7569+%29%29%2F2%5Ca.

n%5B1%5D+=+%28-%283%29%2Bsqrt%28+7569+%29%29%2F2%5C2+=+21
n%5B2%5D+=+%28-%283%29-sqrt%28+7569+%29%29%2F2%5C2+=+-22.5

Quadratic expression 2n%5E2%2B3n%2B-945 can be factored:
2n%5E2%2B3n%2B-945+=+2%28n-21%29%2A%28n--22.5%29
Again, the answer is: 21, -22.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B3%2Ax%2B-945+%29

The series have 21 terms because n cannot be a fraction