Question

ARITHMETIC SERIES WORD PROBLEM!! PLEASE HELP!! \'O'/

Answer 1

Sally just started working at this new daycare center that opened. The first day 100 children came. For each following day for a month, 30 more start coming. How many children come on the 7th day?

Answer 2

Okay, so I know I use the formula Sn = n/2[2a + (n-1)d ]

Answer 3

I'm just a little stuck with what to plug in where...

Answer 4

Sn = 7/2 is all i know so far...

Answer 5

how do i find the common difference?

Answer 6

What is n here?

Answer 7

100?

Answer 8

No, can you tell me what does n symbolizes?

Answer 9

I mean , what does n stand for?

Answer 10

the first term in the series?

Answer 11

wait no! hold on

Answer 12

No! @helpmepassalgebra2

the first term in the series = \(\large a\)

the total terms = \(\large n \)

the last term = \(\large a_n\)

Answer 13

its the position number of a given term

Answer 14

And the common difference = d.

Answer 15

oh ok

Answer 16

so there's 7 terms?

Answer 17

Here I can write the series as :

100, 130, ... , \(a_7\)

as, here , n = 7 so I wrote the last term as \(a_n = a_7\) ... Understanding ?

Answer 18

yes

Answer 19

So now, what is the common difference , @helpmepassalgebra2 ?

Answer 20

uh minus 30?

Answer 21

i mean plus

Answer 22

Good.

Answer 23

okay so that means...Sn = 7/2[2a + (7-1)30 ]

Answer 24

but whats 2a again?

Answer 25

Now, see we have :

n =7

a = 100

d = 30


since : \(\large S_n = \cfrac{n}{2} (2a + (n-1)d) \)

\(\large S_n = \cfrac{7}{2} (200 + 180) \)

\(\large S_n = \cfrac{7}{2} * 380 \)

\(\large S_n = 1330 \)

Answer 26

is it 100? the first term?

Answer 27

2a = 2*a

as : a = 100

so 2a = 200

Answer 28

wow thanks for writing it out like that so i can see how to do it! :D

Answer 29

thank you so much!!

Answer 30

Though, the question is not yet complete my dear friend.

Answer 31

oh really?? O.o what else do we have to do then?

Answer 32

We have to calculate \(a_7\)

And as per the formula :

\(\large S_n = \cfrac{n}{2} (2a + (n-1)d) \)

\(\large S_n = \cfrac{n}{2} ( a +\{ a+ (n-1)d \} )\)

\(\large S_n = \cfrac{n}{2} (a + a_n ) \) \(...\) \(\large \textbf{as :} \space a + (n-1)d = a_n \)

\(1330 = \cfrac{7}{2} ( 100 + a_n) \)

Answer 33

Note, we have to calculate the 7th term , and what we found was "the sum of all terms (S_n) "

Answer 34

okay

Answer 35

Can you calculate it , by understanding the above mentioned method?

Answer 36

what does an respresent again?

Answer 37

sorry idk how to write that like u did.. its a with a small n beside it or under it i guess

Answer 38

a_n represents the last term , here it is 7th term :

\(\large a_7 = \textbf{means, 7th term}\)

Answer 39

okay thx

Answer 40

so 7/2 (100 + 7) = 374.5??

Answer 41

am i write @mathslover ?? or no?

Answer 42

right*

Answer 43

No! you don't have :a_7 = 7

see : let the 7th term be : x

so i have :\(1330 = \cfrac{7}{2} (100 + x) \)

can u solve for x now?

Answer 44

im confused :S

Answer 45

See, we have to calculate the 7th term, right?

Answer 46

yea

Answer 47

Now, I have :

\(\large S_n = \cfrac{n}{2} ( a + a_n) \)

where S_n = sum of all terms = 1330 (we have calculated it already)

a = first term = 100 (given)
a_n = nth term = (here n = 7 ) , so a_n = a_7 = 7th term

thus , I have : \(\large 1330 = \cfrac{7}{2} ( 100 + a_7) \)

Now solve for a_7

Answer 48

so i have to divide 7 and 2 and multiply by 1oo to find a7?

Answer 49

No, see here it is how it goes :

\(\large 1330 = \cfrac{7}{2} ( 100 + a_7) \)

\(\large 1330 * 2 = 7 (100 +a_7) \space ... \space \textbf{I just multiplied 2 both sides } \)

\(\large 2660 = 7(100+a_7)\)

\(\large \cfrac{2660}{7} = 100 + a_7 \space ... \space \textbf{Divided 7 both sides}\)

\(\large 380 = 100 + a_7 \space ... \space \textbf{as 2660/7 = 380 } \)

\(\large{ 380 - 100 = a_7 \space ... \space \textbf{Subtracting 100 both sides}} \)

\(\large 280 = a_7 \space ... \space \textbf{as 380 - 100 = 280} \)

so the required 7th term is 280

Answer 50

woe this was a confusing, LONG process thanks for bearing with me :/