Download ARITHMETIC SERIES

ARITHMETIC
SERIES
Prepared by:
Grace Anne Buno
Michelle Ann Gesmundo
Marty Lordgino Pulutan
What are arithmetic series?
A "series" is the value you get when you add up
all the terms of a sequence; this value is called the
"sum". For instance, "1, 2, 3, 4" is a sequence, with
terms "1", "2", "3", and "4"; the corresponding series
is the sum "1 + 2 + 3 + 4", and the value of the
series is 10.
A series such as 3 + 7 + 11 + 15 + ··· + 99 or 10 + 20 + 30
+ ··· + 1000 which has a constant difference between
terms. The first term is a1, the common difference is d, and
the number of terms is n. The sum of an arithmetic series is
found by multiplying the number of terms times the
average of the first and last terms.
Formula: Sn=n/2(t1+tn) or n/2[2t1+(n-1)d]
wherein the Sn=the sum of n terms
n=the numbers of terms
t1=the first term
tn=the last term
d=the common difference
For getting the n:
n=tn-t1 +1
d
Examples:
1.)1+2+3,…100
let t1 be 1 since the first term is 1.
let n be 100 since there are 100 terms.
let tn be 100 since the nth term is 100.
since all of the terms are needed are given, we
use the formula Sn=n/2(t1+tn).
substitute the values and we have
S100=100/2(1+100).
=5050
therefore, the answer
is
5050
there are other formulas that you could use as well.
• To find tn ,d and t1 tn= t1+(n-1)d
• To find n tn-t1 + 1
d
Let’s try this problem:
if the first term is 14 and the common difference is 2,
find t7 and S7.
The first step is to find the seventh term. Use the
formula tn= t1+(n-1)d.
It will become t7= 14+(7-1)2. the n here is 7.
= 26
Now that we have the 7th term, let us now compute for
the seventh term. Use the formula Sn=n/2(t1+tn).
Again, substitute the values to the formula.
S7= 7/2(14+26)
= 140
therefore:
t7=26 and S7=140
How about this one:
The first term of an A.P. is four and the fourth
term is fourteen find d, t8 and S8.
t1=4
t4=14
D=?
T8=?
S8=?
first, we will look for the d. use the
formula tn= t1+(n-1)d. the n here
would be four.
14=4+(4-1)d
14=4+3d
10/3=d
now that we have our d which is 10/3, we can now
solve for our t8.
Use the formula tn= t1+(n-1)d
t8=4+(8-1)10/3
=4+70/3
=27 and 1/3
Next, we solve for the S8 using the formula
Sn=n/2(t1+tn) .
S8=8/2(4+82/3)
=4(691/3)
=277 1/3
ANSWERS
THE COMMON DIFFERENCE IS 10/3,
THE 8TH TERM IS 27 1/3 AND THE S8 IS
277 1/3.
TAKE THE QUEST
TAKE THE EASY TEST
Now that you know all the steps, try
to complete this table.
t1
d
tn
n
sn
2
5
?
10
?
7
-2
-15
?
?
?
-1.5
15
20
?
THE QUEST
AT THIS PART, YOU WILL BE
GIVEN A SET OF QUESTIONS
FOR YOU TO ANSWER. SOME
QUESTIONS WOULD BE
HARD, EASY OR JUST
AVERAGE. DO YOUR BEST!!!
CONTINUE?
REVIEW AGAIN
PLEASE HELP ME!!!
THE PRINCESS WAS
KIDNAPPED BY THE
ABDUCTORS . . .
SAVE HER . . PLEASE
If you are unable to
do them correctly,
something bad will
happen to your
princess and you
will stay here
forever with me
Solve this problem…WAHAHAHAAAAAAA….!
If the first term is eight and the third term is three, find
t10 and S10.
GOODLUCK. . . . . . . .
TAKE THE NEXT TASK
GIVE UP
I WILL GIVE YOU A
PROBLEM AND IF YOU FAIL
TO GIVE ME THE RIGHT
ANSWER, YOU WILL BE MY
SLAVE FOREVER.AND I
WILL TORTURE YOUR
PRINCESS.
HERE IS YOUR QUESTION:
IF YOUR FOURTH TERM IS 28 AND THE 21TH
TERM IS 100, THEN WHAT IS T15 AND S15?
SOUNDS TOO EASY FOR YOU RIGHT?
TAKE THE LAST CHALLENGE
GIVE UP
ARE YOU READY TO TAKE ON
THE LAST CHALLENGE???
I
THEN WILL GIVE YOU
THE FINAL TASK
IF YOU WOULD BE ABLE TO
SOLVE THIS, THEN YOU COULD
FREELY GO WITH YOUR
PRECIOUS PRINCESS
FIND THE SUM OF ALL POSITIVE
INTEGERS IS LESS THAN 300
WHICH ARE DIVISIBLE BY
NIETHER FIVE OR ELEVEN
END QUEST
YOU HAVE SAVED THE
PRINCESS!!!!!!