Math Homework.. HELP!!!!?

Question

Chapter : Finite and Infinite Sequences( Arithemetic and Geometric Progression)
I need the solved solutions of these two math Questions!
Q1. a1,a2,a3,a4,a5 are first five terms of an Arithemetic progression, such that a1+a3+a5 = -12 and a1*a2*a3=8. Find the first term (T1) and the common difference.
Q2. A side if an equilateral triangle is 20cm long. A second equilateral triangle is inscribed in it by joining the mid points of the sides of the first triangle. The process is continued till the fourth triangle is formed. Find the perimeter of the fourth inscribed equilateral triangle.
The correct answers Ans1= T1=2; Common difference = -3 Ans2= 7.5cm ( as you can see i know the final answers to these questions.. i need the solutions!)
These are grade 11 math questions

Answer 1

an arithmetic progression is a sequence of numbers where the difference between any two successive numbers of the sequence is a constant

so we have as the sequence: a1, a2, a3, a4, a5, etc.
now since it is an arithmetic progression
a2 - a1 = some constant, let's call it c
a3 - a2 = c
a4 - a3 = c
etc.

you know that a1 + a3 + a5 = -12
and a1*a2*a3 = 8
and you're asked to find a1 and the value of c

using the first equation:
a1 + a3 + (a5-a4) +a4 = -12
(I added and subtracted a4 on left side)
but note that a5-a4 = c, so we can rewrite it as
the difference of any two sucessive terms which would
be convenient to solving the problem
let's use a2-a1
a1 + a3 + (a2-a1) +a4 = -12
let us do the same trick for the a3 and a4 terms
a1 + (a3 - a2 + a2) + (a2 -a1) + (a4 - a3 + a3) = -12
a1 + (a2 - a1 + a2) + (a2 -a1) + (a2 - a1 + a3) = -12
now collect like terms
-2a1 + 4a2 + a3 = -12
let's repeat that trick to get rid of a3
-2a1 + 4a2 + (a3 - a2 + a2) = -12
-2a1 + 4a2 + a2-a1 +a2 = -12
-3a1 + 6a2 = -12
let's divide by 3
-a1 + 2a2 = -4
rewrite this as
a2 + (a2 - a1) = -4
this tells you a3 = -4
[because we could rewrite (a2 - a1) as (a3 - a2)]
now that we know a3, we can put this value in the 2nd equation
and get
a1*a2= 8/-4 = -2
or a2 = -2/a1
substitute this into our earlier equation of
2a2 - a1 = -4
to get
2(-2/a1) -a1 = -4
multiply by a1
-4 -a1^2 = -4a1
rearrange
a1^2 - 4a1 + 4 = 0
factor
(a1 - 2)^2 = 0
a1 = 2
now you know a1 which is T1 = 2
go back to 2a2 - a1 = -4
2a2 -2 = -4
a2 = -1
then common difference is
a2 - a1 = -1 - 2 = -3
I might have been able to make the solution a little more direct.
nevertheless i hope you understand the solution and can work it out yourself.

for Q2, the side of each inscribed triangle will be half that of the side of the triangle that it is inscribed in (draw it and it should be obvious)
so the length of a side of the 4th triangle is
20 cm / 2 / 2 / 2 = 2.5 cm
the perimeter is just 3 times this (there are 3 sides to a triangle)
3 * 2.5 cm = 7.5 cm

Answer 2

Q1.
3a1 +2d + 4d = -12
3a1 + 6d = -12
a1 + 2d = -4
a1 = -2(d + 2)
a1(a1+d)(a1+2d) = 8
-2(d + 2)(-2(d + 2) + d)(-2(d + 2) + 2d) = 8
(d + 2)(d + 4) = - 1
d^2 + 6d + 9 = 0
(d + 3)^2 = 0
d = ± 3
a1 = -2(d + 2) = -10,2
-10*-7*-4 ≠ 8
2*-1*-4 = 8
d = -3
a1 = 2

Q2.
L1 = 20
L2 = L1/2 (forms new equilateral triangle)
L3 = L1/4
L4 = L1/8 = 20/8 =2.5
P4 = 3*2.5 = 7.5

Answer 3

go ahead and find the solution yourself my friend.you have found out the answer ..now work it out patiently with the formula given inyour text book.. online friends wont be able to help you in your final exams. so pick up your math text and start working on them right a way..
goodluck

Answer 4

You shouldn't be using ppl online to do ur math homework.
You can use tutor.com.. for instance.. if u need help..
getting solutions this way won't solve ur upcoming test problems later on.