word problems sequences and series

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Please help I have ALOT of questions!!

1.)Starting at 888 and counting backward by 7, a student counts 888, 881 and 874, and so on. Which of the following numbers will be included?

a) 35 b) 34 c) 33 d) 32 e) 31

Ok, so I by using the calcuator the aswer is 34, but how would you calcuate this using an equation or something more sensible...

I think I have to use this one

tn=a + (n-1) d

a= 888
n=
d= -7
tn=

... now what

__________________________________________________ ___________

2. the sum of the first n even postive integers is h. the sum of the first n odd positive integers is k. then h -k is equal to:

a) n/2
b) n/2 -1
c) n
d) -n
e) n - 1

so... nK and nh

I don't get it...

__________________________________________________ __________

3. the largest four- digit number to be foung in the arithmetic sequence
1,6,11,16,21, ... is

t1= 1
t2= 6
t3= 11
t4= 21

9996?
because it goes form 1 to 6 the (last number)
2nd last number stayes for 2 ed 1,1,2,2
everything else up one

but is there an equation for this or something?
__________________________________________________ _________
4. the sum of 50 consecutive intergers is 3250. tha largest of these intergers is

a)64
b)66
c)112
d)114
e)115

I don't know how to so this...is it the same as before?
__________________________________________________ _________
5 Jan 1, 1986 was a wed. Jan 1 1992 was waht day of the week?

86 wed
87 thurs
88 fri
89 sat
90 sun
91 mon
92 tues

tuesday?

what would an equatin be
__________________________________________________ _________

6) The total number of digits used to number all the pages of a book was 216. Find the number of pages in the book.

tn= a + (n-1) d
216= 1 + (n- 1) 1
216= 1 + n - 1
216=n

is that wrong...

so...
total number of digits? sum?

Sn= n/2 [ 2a + ( n-1) d ]
216= n/2 [ 2(1) + (1-1) 1]
216= n/2 (2)
216= n

umm....?

Pleas help!
thanks
 
Aka said:
2. the sum of the first n even postive integers is h. the sum of the first n odd positive integers is k. then h -k is equal to:

a) n/2
b) n/2 -1
c) n
d) -n
e) n - 1

so... nK and nh

I don't get it...
I'll give you a hand with this one. It's getting late.(yawn)

\(\displaystyle \text{Use the arithmetic series thing\\

For the positive:}\)

\(\displaystyle \L\\S_{n}=\frac{n}{2}[2a_{1}+(n-1)d]\)

\(\displaystyle a_{1}=2\\d=2\)

\(\displaystyle \L\\h=\frac{n}{2}[2(2)+(n-1)(2)]\)

\(\displaystyle \text{odds would be:}\) \(\displaystyle \L\\k=\frac{n}{2}[2(1)+(n-1)(2)]\)

\(\displaystyle \text{Now subtract h and k. Your answer is there among the choices.}\)
 
Hello, Aka!

Here are a few of them . . .

1) Starting at 888 and counting backward by 7, a student counts 888, 881 and 874, and so on.
Which of the following numbers will be included?

\(\displaystyle a)\;35\qquad\qquad b)\;34\qquad\qquad c)\:33\qquad\qquad d)\;32\qquad\qquad e)\;31\)
Your reasoning is correct . . .
We'll use: \(\displaystyle \,t_n\:=\:t_1\,+\,d(n\,-\,1)\)

We have: \(\displaystyle \,t_1\,=\,888,\;\;d\,=\,-7\)

So: \(\displaystyle \,t_n\:=\:888\,-\,7(n\,-\,1)\)

Solve for \(\displaystyle n:\;\;n\:=\:\frac{888\,-\,t_n}{7}\,+\,1\)

Since \(\displaystyle n\) is positive integer, \(\displaystyle 888\,-\,t_n\) must be divisible by 7.

And only \(\displaystyle t_n\,=\,34\) works . . . answer (b)


2. The sum of the first \(\displaystyle n\) even postive integers is \(\displaystyle h\).
The sum of the first \(\displaystyle n\) odd positive integers is \(\displaystyle k\).
Then \(\displaystyle h\,-\,k\) is equal to:

\(\displaystyle a)\;\frac{n}{2}\qquad\qquad b)\;\frac{n}{2}\,-\,1\qquad\qquad c)\;n\qquad\qquad d)\;-n\qquad\qquad e)\;n\;-\,1\)
I'll solve this from square-one . . .


The sum of the first \(\displaystyle n\) even integers: \(\displaystyle \,2\,+\,4\,+\,6\,+\,\cdots\,+\,2n\)

\(\displaystyle \;\;\) is an arithmetic series with \(\displaystyle t_1\,=\,2\) and \(\displaystyle d\,=\,2\)

Hence: \(\displaystyle \,h\;=\;S_{\text{even}}\;=\;\frac{n}{2}[2\cdot2\,+\,2(n-1)] \;=\;n(n\,+\,1)\)


The sum of the first \(\displaystyle n\) odd integers: \(\displaystyle \,1\,+\,3\,+\,5\,+\,+\,\cdots\,+\,(2n-1)\)

\(\displaystyle \;\;\)is an arithmetic series with \(\displaystyle t_1\,=\,1\) and \(\displaystyle d\,=\,2\)

Hence: \(\displaystyle \,k\;=\;S_{\text{odd}}\;=\;\frac{n}{2}[2\cdot1\,+\,2(n-1)] \;= \;n^2\)


Therefore: \(\displaystyle \,h\,-\,k\;=\;n(n\,+\,1)\,-\,n^2\;=\;n\) . . . answer (c)


3. the largest four-digit number to be foung in the arithmetic sequence
1,6,11,16,21, ... is >
Your answer (9996) is correct!

We have an arithmetic sequence with \(\displaystyle t_1\,=\,1,\;d\,=\,5\)

The \(\displaystyle n^{th}\) term is: \(\displaystyle \,t_n\;=\;1\,+\,5(n\,-\,1)\)

We want \(\displaystyle t_n\) to be less than 10,000 (a five-digit number).

So we have: \(\displaystyle \,1\,+\,5(n\,-\,1)\;<\;10,000\)

\(\displaystyle \;\;\)Subtract 1: \(\displaystyle \,5(n\,-\,1)\;<\;9,999\)

\(\displaystyle \;\;\)Divide by 5: \(\displaystyle \,n\,-\,1\;<\;1999.8\)

\(\displaystyle \;\;\)Add 1: \(\displaystyle \,n \;<\;2000.8\)

So, we let \(\displaystyle \,n\,=\,2000\)

Then: \(\displaystyle t_{_{2000}}\;=\;1\,+\,5(1999)\;=\;9996\) . . . There!
 
Aka said:
1.)Starting at 888 and counting backward by 7, a student counts 888, 881 and 874, and so on. Which of the following numbers will be included?

a) 35 b) 34 c) 33 d) 32 e) 31

Ok, so I by using the calcuator the aswer is 34, but how would you calcuate this using an equation or something more sensible...

I think I have to use this one

tn=a + (n-1) d

a= 888
n=
d= -7
tn=

... now what
Dividing each number of the series by 7 leaves a remainder of 6.

Which of ) 35 b) 34 c) 33 d) 32 e) 31 leaves a remander of 6 when divided by 7?

34/7 = 4 + 6 remainder.
 
Aka said:
2. the sum of the first n even postive integers is h. the sum of the first n odd positive integers is k. then h -k is equal to:

a) n/2
b) n/2 -1
c) n
d) -n
e) n - 1

so... nK and nh

I don't get it...
The sum of the first n even integers is n(n + 1).

The sum of the first n odd integers is n^2.

Therefore, the difference between the first n even integers and the first n odd integers is n(n + 1) - n^2 = n^2 + n - n^2 = n.
 
Aka said:
3. the largest four- digit number to be foung in the arithmetic sequence 1,6,11,16,21, ... is: t1 = 1, t2 = 6, t3 = 11, t4 = 21

9996? because it goes form 1 to 6 the (last number)
2nd last number stayes for 2 ed 1,1,2,2 everything else up one

but is there an equation for this or something?
9999/5 = 1999.8 or 1999 + 4

Therefore, the highest 4 digit number in the series is 1999 - 4 = 1996.

Every number leaves a remainder of 1 when divided by 5.

The number of terms is 1996/5 = 399.2 or 399 with a 1 remainder
 
Hello, Aka!

4. The sum of 50 consecutive intergers is 3250. tha largest of these intergers is:

\(\displaystyle a)\;64\qquad\qquad b)\;66\qquad\qquad c)\;112\qquad\qquad d)\;114\qquad\qquad e)\;115\)
The answer is "None of These"!


We have an arithmetic series: \(\displaystyle a\,+\,(a+1)\,+\,(a+2)\,+\,\cdots\,+\,(a+49)\)

The sum of the first \(\displaystyle n\) terms is: \(\displaystyle \,S_n\;=\;\frac{n}{2}[2a_1\,+\,d(n-1)]\)

Our first term is \(\displaystyle a\), the common difference is \(\displaystyle d = 1,\;n\,=\,50,\;S_{50}\,=\,3250\)

We have: \(\displaystyle \,3250\;=\;\frac{50}{2}[2\cdot a\,+\,1(50\,-\,1)]\)

\(\displaystyle \;\;\)then: \(\displaystyle \,3250\;=\;25[2a\,+\,49)\;\;\Rightarrow\;\;130\;=\;2a\,+\,49\;\;\Rightarrow\;\;81\;=\;2a\)

But this gives us: \(\displaystyle \,a\;=\;40.5\) . . . and \(\displaystyle a\) is supposed to be an integer.

So I assume there is a typo in the statement of the problem.


5. January 1, 1986, was a Wednesday.
January 1, 1992, was what day of the week?
There is no "formula" for this problem; you must do some Thinking.

From 01/01/86 to 01/01/92 is six years: \(\displaystyle \,6\,\times\,365\:=\:2190\) days.
But 1988 was a leap year, so there are: 2191 days.

Since \(\displaystyle 2191\,\div\,7\;=\;313\) with no remainder,
\(\displaystyle \;\;\)then 01/01/92 is exactly 313 weeks after 01/01/86.

Therefore, January 1, 1992 is also on a Wednesday.


6) The total number of digits used to number all the pages of a book was 216.
Find the number of pages in the book.
There is no formula for this one either . . . You must baby-talk your way through it.

Pages 1 to 9: nine 1-digit numbers = 9 digits.

Pages 10 to 99: ninety 2-digit numbers = 180 digits.

There are: \(\displaystyle \,216\,-\,9\,-\,90\:=\:27\) digits to go.

These are taken up by the first nine 3-digit numbers: 100, 101, 102, ... , 108

Therefore, the last page is number 108.
 
Aka said:
4. the sum of 50 consecutive intergers is 3250. tha largest of these intergers is: a) 64, b) 66, c) 112, d) 114, e) 115
Assuming the first and last numbers are x and y,
(x + y)50/2 = 3250 making x +y = 130
For 50 integers, x - y = 51
x + y = 130
x - y = 51
2x = 181 making x 90.5

since we are supposed to be dealing with integers, some given information is wrong.
 
Thanks for everyones help!

Sorry, I typed #4 wrong its

4. the sum of 50 consecutive even intergers is 3250. tha largest of these intergers is

a)64
b)66
c)112
d)114
e)115
 
That makes a world of difference then.

\(\displaystyle \L\\25(2x+98)=3250\)

\(\displaystyle x=16\)

The first number of the series is 16. Can you figure out what the last one is now?.
 
^ yes, thanks

One more question

Tod injured his wrist while he was skateboarding. To relive the pain his doctor prescribed him some medication. He takes 500mg of medicine every 6h. Only 25 % of the medication remains in his body by the time he is ready to take another pill.
Write and equation to model this.


so 25% is left in his body that means that 75% was used up.

so 25%, 125%, !50%

...thats wrong right?

I dont know what to do...
 
galactus said:
That makes a world of difference then.

25(2x+98)=3250

x =16

The first number of the series is 16. Can you figure out what the last one is now?
With a first number of 16:
An upper number of 58 would yield 58(58 + 1) - 15(15 + 1) = 3182, not 3250 using only 22 even numbers.

Either 50 numbers, or 3250 is wrong.

The first 50 even numbers alone add up to 2550.
 
Aka said:
Thanks for everyones help!

Sorry, I typed #4 wrong its

4. the sum of 50 consecutive even intergers is 3250. tha largest of these intergers is

a)64
b)66
c)112
d)114
e)115

The sum of the first 50 even numbers starting with 2 is only 2550. Either your 50 consecutive even numbers or the 3250 is wrong. Check your problem statement again.

A suggested first number of 16 combined with an upper number of 58 yields a total of 58(58+1) - 15(15+1) = 3182 using only 22 even numbers.
 
I don't know, TchrWill, seems OK.

\(\displaystyle \L\\\sum_{n=8}^{57}(2n)=3250\)

That would be 16+18+20+22+................+114=3250.

That's 50 consecutive even integers.
 
galactus said:
I don't know, TchrWill, seems OK.

\(\displaystyle \sum_{n=8}^{57}(2n)=3250\)

That would be 16+18+20+22+................+114=3250.

That's 50 consecutive even integers.
My most humble apologies. 16 through 114 is clearly the correct answer and I now see where my thinking went astray. Sorry for the misdirections along the way.

I should have gone with my first instinct of dividing 3250 by 50 yielding 65 making the sequence of numbers the 25 even numbers on both sides of 65, i.e., 16 through 114. Oh well, will be more careful in the future.
 
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