Solving for n in a Summation Series

[mattyh3]

Homework Statement

How many terms of the AP, 12, 16, 20, 24 . . . . . must be taken for the sum to
equal 408?
Solution
For this series, a = 12, d = 4 and Sn = 408.
its asking me to solve for n and gives me the answer to n as 12 at the end but when i try to calculate it myself i don't get 12 and it only states limited steps and i a bit unsure on how to transpose the equation to make n the subject so it comes out with the answer 12

Homework Equations

sn= n/2 (2a + (n-1)d)
then its has

408=n/2 (2 x 12 +(n-1) x 4)

but am not to great at transposing and can't seem to make n the subject and get the answer of 12 can anyone help with this as its only an example on my lesson book i just want to understand how its transposing to get n = 12

Solving this for n gives n = 12. The other root of the equation is negative and
is rejected as impractical.

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[voko]

I am not sure what your problem is. The root of the quadratic equation is indeed 12, and it is the correct answer.

 

[mattyh3]

yeah i know the answer is 12 because it tells me... but i am not sure how it is coming up with 12 as i am unsure on how to make n the subject so i can work it out myself as i need to understand how for assignment question later on

 

[voko]

You said that you solved the equation and got 12 and some other negative root. This is really all it takes; there is nothing else.

 

[mattyh3]

i haven't solved it the lesson book did the equation and at the end said n=12.. but what am trying to say is i don't know how its transposed the equation to make n the subject and to come up with n=12 i was wondering if someone could fill in the missing steps as to how it arrives with n=12

 

[voko]

Have you studied quadratic equations? Does this: $ax^2 + bx + c = 0$ look familiar?

 

[mattyh3]

i have studied them but many years ago and i have seen that before but that's why am struggle and i can't seem to transpose that equation to find n and i have tried loads of times and i end up with answers way off

 

[voko]

You can google for details, but the solution to the equation given in #6 is $$ x = \frac {-b \pm \sqrt{D}} {2a} $$ where $D = b^2 - 4ac$. You need to get the equation for the sum of progression in the form as in #6 and then solve it.

 

[mattyh3]

i have found on google how to do what is in 6 but looking at that i aint got a clue on where to start with getting n as the subject in this sn=n/2 (2a+(n-1)d)..

any help would be great just to get me started even a first step or so as its only an example in my book and i just want to be able to do it

 

[voko]

'n' plays the role of 'x' - this is the unknown you are to find out.

The first step is to "open the brackets" in you formula, using the distributive law of arithmetic. Then you have to collect all the terms on one side of the equation (so that you have zero on the other side), and then identify the coefficients with those in the #6 formula.

 

[mattyh3]

its knowing/ remembering how to do that as well. am really struggling here :( as mine seems a lot harder to do than what is in step 6

 

[voko]

For example: (n - 1)d = dn - d. Then you need to repeat that for n/2 (2a + dn - d). Nothing fancy here.

 

[mattyh3]

still confused as to what's happeneds to the Sn so to me the next step would be 2a + 2dn-2dn or am i wrong

 

[mattyh3]

still confused as to what's happeneds to the Sn so to me the next step would be 2a + 2dn-2dn or am i wrong,,, its easy when you know how and i don't lol

but am guessing with 2dn-2dn they would cancel each other out

 

[voko]

To simplify things, you can multiply the equation - both sides - by 2. That will make it 2s = n(...).

 

[mattyh3]

or sn= n+ 2a +dn-d/2 then sn= 2a+d2n-d/2 think am going totally wrong as am not to sure what the next step is etc

 

[mattyh3]

so 2Sn= n2a + dn-d am i on the right lines or still off

 

[voko]

Distributive law: A(B + C + D + ... + Z) = AB + AC + AD + ... + AZ.