# Thread: word problems sequences and series

1. ## word problems sequences and series

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

2. ## Re: word problems sequences and series

Originally Posted by Aka
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)

$\text{Use the arithmetic series thing\\ For the positive:}$

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

$a_{1}=2\\d=2$

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

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

$\text{Now subtract h and k. Your answer is there among the choices.}$

3. ## Re: word problems sequences and series

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?

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

We have: $\,t_1\,=\,888,\;\;d\,=\,-7$

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

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

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

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

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:

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

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

$\;\;$ is an arithmetic series with $t_1\,=\,2$ and $d\,=\,2$

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

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

$\;\;$is an arithmetic series with $t_1\,=\,1$ and $d\,=\,2$

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

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

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

We have an arithmetic sequence with $t_1\,=\,1,\;d\,=\,5$

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

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

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

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

$\;\;$Divide by 5: $\,n\,-\,1\;<\;1999.8$

$\;\;$Add 1: $\,n \;<\;2000.8$

So, we let $\,n\,=\,2000$

Then: $t_{_{2000}}\;=\;1\,+\,5(1999)\;=\;9996$ . . . There!

4. ## Re: word problems sequences and series

Originally Posted by Aka
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.

5. ## Re: word problems sequences and series

Originally Posted by Aka
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.

6. ## Re: word problems sequences and series

Originally Posted by Aka
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

7. ## Re: word problems sequences and series

Hello, Aka!

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

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

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

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

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

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

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

But this gives us: $\,a\;=\;40.5$ . . . and $a$ 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: $\,6\,\times\,365\:=\:2190$ days.
But 1988 was a leap year, so there are: 2191 days.

Since $2191\,\div\,7\;=\;313$ with no remainder,
$\;\;$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: $\,216\,-\,9\,-\,90\:=\:27$ 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.

8. ## Re: word problems sequences and series

Originally Posted by Aka
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.

9. 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

10. That makes a world of difference then.

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

$x=16$

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

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