Warm Up Questions

2nd question:

"2. The mean of the first five terms of an arithmetic sequence is 27, and the mean of the first eight terms of the same sequence is 39. What is the absolute difference between the first and second terms of this sequence?" By the way, what do you mean by the absolute difference between the first & second terms of this sequence? so we subtract the first mean from the second mean? That's what I did.. or we subtract the first term from the second term?

I think:

Let the first 5 terms of the arithmetic sequence be:   X , Y , Z , K, L

We know:

(X+Y+Z+K+L) /5 =27

"The mean of the first eight terms of the same sequence is 39" So I understood that it is the extention of the first mean 

Let the new 3 terms be F,S,Q

(X+Y+Z+K+L)/5 + F+S+Q=39 

we know that (X+Y+Z+K+L)/5=27

27+F+S+Q=39

F+S+Q=12

Ok so "Difference between the first and second terms of this sequence" 

first is 27-12=15

Not so sure.. maybe I understood it wrong.

2)   5/2 * [2F + D(5 - 1)] =135.............(1)

       8/2 * [2F + D(8 - 1) ] =312...........(2), solve for F, D

F = 11 - which is the first term and D = 8 - which is the common difference.

The sequence is: 11, 19, 27, 35, 43, 51, 59, 67 . The mean of the first 5 terms is 27 and of all 8 terms is 39. The absolute difference between the first and the second is: abs[11 - 19] = 8, Or: 19 - 11 =8.

Hi CalculatorUser,

Question 1

I have not answered this question. I have not even tried yet

I did draw it with geogebra, because that looked fun.     

This is the graphical approximation.

Question 1

I do not know how to do question one but I'd like to see someone else's solution.

It has probably got something to do with similarity....  maybe ...    

2. The mean of the first five terms of an arithmetic sequence is 27, and the mean of the first eight terms of the same sequence is 39. What is the absolute difference between the first and second terms of this sequence?

Let the first five terms be  

a , (a + d), (a + 2d), (a + 3d), (a + 4d)..where "a" is the first term and "d"  is the common difference between terms....the sum of these  =  5a + 10d

So....

(5a + 10d)

________  =   27     →     a + 2d  =  27        (1)

      5

And the next three terms  are      (a + 5d), (a + 6d), (a + 7d)

So......the sum of the first 8 terms  =   8a + 28d

So.....

(8a + 28d)

_______      =  39   →     a + (7/2)d  =  39        (2)

      8

Subtract  (1)  from (2)    and we get that    (3/2)d   =  12    →    d = 8

So...the absolute difference between the first and second terms is just "d"    =  l  8  l   =   8

1. 

Here's a pic : 

 Call the intersection of BC and  B'A' , N

Note that triangle BMN   is similar to triangle BCA

So  

MN/ BM  = AC / BC

MN / (5/2)  = 3/4

MN  = 15/8

So....the area of triangle BMN  =  (1/2) (product of the legs) = (1/2)(BM)(MN) = (1/2)(5/2)(15/8)  = 75/32

So BN  = sqrt [ BM^2 + MN^2]  = sqrt [ (5/2)^2 +(15/8)^2 ] =  25/8 

So B'N  =   B'M - MN =   (5/2) - (15/8)  =  20/8 - 15/8 =  5/8

So  A'N  = A'M + MN  =   5/2  + 15/8  =   20/8 + 15/8 = 35/8

And call the intersection of BC  and B'C', P

Because BC  is parallel to A'C'.....then

A'N / A'B'   = C'P / C'B'

(35/8) /  5  =  C'P /4

35/40  = C'P /4

C'P  = 3.5

Then B'P  = B'C' - C'P  =  4 -3.5  = .5

So....using the Pythagorean Theorem....NP  =  sqrt [B'N^2 - B'P^2] =  sqrt [ (5/8)^2 - (4/8)^2 ]  = 3/8

Then  BP  =  BN + NP  =  25/8 + 3/8  =  28/8 =  7/2

And call the intersection of BA and B'C', Q

And because PQ  is parallel to AC

Then     BP / BC  = PQ/CA

(7/2) / 4 =  PQ/ 3

(7/8) = PQ/3

(21/8)  = PQ

So  the area of triangle BPQ  =  (1/2) (BP)(PQ)  = (1/2) (7/2) (21/8)  =  147/32

So... the shaded area  = [area of triangle BPQ ]  - [area of triangle BMN ] =

[ 147 - 75 ] / 32  =  72/32 =  9/4  =  2.25 units^2

Thanks CPhill, Melody, and Guest. I think Triangle similarity is something I need to work on more!