Solving for x in an Arithmetic Progression with Given Sum and Terms

  • Thread starter Thread starter mark-ashleigh
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around solving for the variable x in an arithmetic progression where the first term is expressed as (1-x)^2 and the second term as 1+x^2. The sum of the first ten terms is given as 310, and participants are exploring how to derive the possible values of x.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss expanding the expressions for the first two terms and using quadratic equations to find values for x. There are questions about the correctness of the sum and the signs of the differences between terms. Some participants suggest writing the difference between the first two terms as a function of x and using it to derive further terms.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem. There are multiple interpretations being explored, particularly regarding the signs of the differences and the setup of the sum formula. Some participants express confusion about the expressions and calculations, indicating a need for clarification.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There are indications of misreading expressions and questioning the validity of the problem setup.

mark-ashleigh
Messages
30
Reaction score
0
The first term of an arithmetic progression is (1-x)^2 and the second term is 1+x^2 .If the sumj of the first ten terms is 310 , find the possible values of x.

I have my A/S maths exam next month, but i am still having trouble with arithmetic progression. The above question is causing me some trouble .

First i expanded the brackets using binomial expansion .

Then as i had a quadratic i used the theorem to find values for x .

Once i found x i substituted into the first two terms to find the difference .

I found the first term = 1.98 the second = 6.8 with a diff of 4.8 .

As a + ( 9 X d ) = the tenth term = 45.36

And the formula for the sum is

S 10 = 10 x ( a + l)/2 ...where l = 45.36

Why do i keep getting 236 .7

Am i doing something drasticaly wrong?

Many thanks .
 
Physics news on Phys.org
You cannot find numerical values for the first two terms until you find all the terms needed to satisfy the sum. Write the difference between the first two terms as a function of x, and use that to write the difference between the first and last term. Then use what you know about how to sum an arithmetic progression.
 
is the difference -2x?
 
even with the common factor the sum doesn't = 310 .

Is the question incorrect?
 
mark-ashleigh said:
even with the common factor the sum doesn't = 310 .

Is the question incorrect?

Sorry. I may have misread the expressions.

[tex](1 - x)^2 = 1 - 2x + x^2[/tex]

[tex]1 + x^2 - (1 - x)^2 = 1 + x^2 - 1 + 2x - x^2 = 2x[/tex]

Is this what you have?
 
Last edited:
Yes, but i got -2x

if this is the difference can i substitute it into

(1-x)^2 + ( 9 x - 2x) = 10 the term

then solve for x before sub into

S10 = 10 x (A + L) / 2

?
Thanks
 
mark-ashleigh said:
Yes, but i got -2x

if this is the difference can i substitute it into

(1-x)^2 + ( 9 x - 2x) = 10 the term

then solve for x before sub into

S10 = 10 x (A + L) / 2

?
Thanks

the notation is confusing. I think you mean

(1-x)^2 + (9)(- 2x) = the 10th term

but the difference is the second term minus the first term. Get the sign of the difference right. Then substitute that expression for the 10th term into your expression for the sum and set the sum = 310. Then solve for x.
 
Thank you...
 
i was going down the right track...but must have lost myself with the wrong sign...
thanks :)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
4
Views
2K