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ARITHMETIC SERIES Prepared by: Grace Anne Buno Michelle Ann Gesmundo Marty Lordgino Pulutan

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ARITHMETIC SERIES. Prepared by: Grace Anne Buno Michelle Ann Gesmundo Marty Lordgino Pulutan. What are arithmetic series?. - PowerPoint PPT Presentation

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Page 1: ARITHMETIC SERIES

ARITHMETIC SERIES

Prepared by:Grace Anne Buno

Michelle Ann GesmundoMarty Lordgino Pulutan

Page 2: ARITHMETIC SERIES

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.

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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

Page 4: ARITHMETIC SERIES

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 haveS100=100/2(1+100).

=5050 therefore, the answer is 5050

Page 5: ARITHMETIC SERIES

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

dLet’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).

Page 6: ARITHMETIC SERIES

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 first, we will look for the d. use the

t4=14 formula tn= t1+(n-1)d. the n here

D=? would be four.T8=?

S8=?

Page 7: ARITHMETIC SERIES

14=4+(4-1)d14=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) .

Page 8: ARITHMETIC SERIES

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.

Page 9: ARITHMETIC SERIES

TAKE THE QUEST

TAKE THE EASY TEST

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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 ?

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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

Page 13: ARITHMETIC SERIES

PLEASE HELP ME!!! THE PRINCESS WAS KIDNAPPED BY THE

ABDUCTORS . . .SAVE HER . . PLEASE

Page 14: ARITHMETIC SERIES

If you are unable to do them correctly,

something bad will happen to your

princess and you will stay here

forever with me

Page 15: ARITHMETIC SERIES

Solve this problem…WAHAHAHAAAAAAA….!

If the first term is eight and the third term is three, find t10 and S10.

GOODLUCK. . . . . . . .

TAKE THE NEXT TASKGIVE UP

Page 16: ARITHMETIC SERIES

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.

Page 17: ARITHMETIC SERIES

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 CHALLENGEGIVE UP

Page 18: ARITHMETIC SERIES

ARE YOU READY TO TAKE ON THE LAST CHALLENGE???

THEN I WILL GIVE YOU THE FINAL TASK

Page 19: ARITHMETIC SERIES

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

Page 20: ARITHMETIC SERIES

YOU HAVE SAVED THE PRINCESS!!!!!!

Page 21: ARITHMETIC SERIES