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

2. 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.

3. 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. seriesconstantdifference termssum average Formula: S n =n/2(t 1 +t n ) or n/2[2t 1 +(n-1)d] wherein the S n =the sum of n terms n=the numbers of terms t 1 =the first term t n =the last term d=the common difference For getting the n: n=t n -t 1 +1 d

4. Examples: 1.)1+2+3,…100 let t 1 be 1 since the first term is 1. let n be 100 since there are 100 terms. let t n be 100 since the n th term is 100. since all of the terms are needed are given, we use the formula S n =n/2(t 1 +t n ). substitute the values and we have S 100 =100/2(1+100). =5050therefore, the answer is 5050

5. there are other formulas that you could use as well. To find t n,d and t 1  t n = t 1 +(n-1)d To find n  t n -t 1 + 1 d Let’s try this problem: if the first term is 14 and the common difference is 2, find t 7 and S 7. The first step is to find the seventh term. Use the formula t n = t 1 +(n-1)d. It will become t 7 = 14+(7-1)2. the n here is 7. = 26 Now that we have the 7 th term, let us now compute for the seventh term. Use the formula S n =n/2(t 1 +t n ).

6. Again, substitute the values to the formula. S 7 = 7/2(14+26) = 140therefore: t 7 =26 and S 7 =140 How about this one: The first term of an A.P. is four and the fourth term is fourteen find d, t 8 and S 8. t 1 =4first, we will look for the d. use the t 4 =14formula t n = t 1 +(n-1)d. the n here D=? would be four. T 8 =? S 8 =?

7. 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 t 8. Use the formula t n = t 1 +(n-1)d t 8 =4+(8-1)10/3 =4+70/3 =27 and 1/3 Next, we solve for the S 8 using the formula S n =n/2(t 1 +t n ).

8. S 8 =8/2(4+82/3) =4(691/3) =277 1/3 ANSWERS THE COMMON DIFFERENCE IS 10/3, THE 8 TH TERM IS 27 1/3 AND THE S 8 IS 277 1/3.

9. TAKE THE QUEST TAKE THE EASY TEST

11. Now that you know all the steps, try to complete this table. t1t1 dtntn nsnsn 25?10? 7-2-15?? ?-1.51520?

12. 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

13. PLEASE HELP ME!!! THE PRINCESS WAS KIDNAPPED BY THE ABDUCTORS... SAVE HER.. PLEASE

14. If you are unable to do them correctly, something bad will happen to your princess and you will stay here forever with me

15. Solve this problem…WAHAHAHAAAAAAA….! If the first term is eight and the third term is three, find t 10 and S 10. GOODLUCK........ TAKE THE NEXT TASK GIVE UP

16. 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.

17. HERE IS YOUR QUESTION: IF YOUR FOURTH TERM IS 28 AND THE 21TH TERM IS 100, THEN WHAT IS T 15 AND S 15 ? SOUNDS TOO EASY FOR YOU RIGHT? TAKE THE LAST CHALLENGE GIVE UP

18. ARE YOU READY TO TAKE ON THE LAST CHALLENGE??? THEN I WILL GIVE YOU THE FINAL TASK

19. 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

20. YOU HAVE SAVED THE PRINCESS!!!!!!