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

- anonymous

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

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- anonymous

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?

- anonymous

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

- anonymous

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

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## More answers

- anonymous

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

- anonymous

how do i find the common difference?

- mathslover

What is n here?

- anonymous

100?

- mathslover

No, can you tell me what does n symbolizes?

- mathslover

I mean , what does n stand for?

- anonymous

the first term in the series?

- anonymous

wait no! hold on

- mathslover

No! @helpmepassalgebra2
the first term in the series = \(\large a\)
the total terms = \(\large n \)
the last term = \(\large a_n\)

- anonymous

its the position number of a given term

- mathslover

And the common difference = d.

- anonymous

oh ok

- anonymous

so there's 7 terms?

- mathslover

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 ?

- anonymous

yes

- mathslover

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

- anonymous

uh minus 30?

- anonymous

i mean plus

- mathslover

Good.

- anonymous

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

- anonymous

but whats 2a again?

- mathslover

Now, see we have :
n =7
a = 100
d = 30
since : \(\large S_n = \cfrac{n}{2} (2a + (n-1)d) \)
\(\large S_n = \cfrac{7}{2} (200 + 180) \)
\(\large S_n = \cfrac{7}{2} * 380 \)
\(\large S_n = 1330 \)

- anonymous

is it 100? the first term?

- mathslover

2a = 2*a
as : a = 100
so 2a = 200

- anonymous

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

- anonymous

thank you so much!!

- mathslover

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

- anonymous

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

- mathslover

We have to calculate \(a_7\)
And as per the formula :
\(\large S_n = \cfrac{n}{2} (2a + (n-1)d) \)
\(\large S_n = \cfrac{n}{2} ( a +\{ a+ (n-1)d \} )\)
\(\large S_n = \cfrac{n}{2} (a + a_n ) \) \(...\) \(\large \textbf{as :} \space a + (n-1)d = a_n \)
\(1330 = \cfrac{7}{2} ( 100 + a_n) \)

- mathslover

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

- anonymous

okay

- mathslover

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

- anonymous

what does an respresent again?

- anonymous

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

- mathslover

a_n represents the last term , here it is 7th term :
\(\large a_7 = \textbf{means, 7th term}\)

- anonymous

okay thx

- anonymous

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

- anonymous

am i write @mathslover ?? or no?

- anonymous

right*

- mathslover

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?

- anonymous

im confused :S

- mathslover

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

- anonymous

yea

- mathslover

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

- anonymous

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

- mathslover

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

- anonymous

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

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