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 constantdifference 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 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=?

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.

challenges TAKE THE QUEST TAKE THE EASY TEST

EXERCISES

Now that you know all the steps, try to complete this table. TAKE THE QUEST FINISHED

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??? THEN I 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!!!!!! CONGRATULATIONS

THANK YOU!!! THE END

ARITHMETIC SERIES