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

  • Thread starter mark-ashleigh
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In summary, when graphing an arithmetic progression, the first term is (1-x)^2 and the second term is 1+x^2. If the sum of the first ten terms is 310, find the possible values of x.
  • #1
mark-ashleigh
30
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 .
 
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  • #2
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.
 
  • #3
is the difference -2x?
 
  • #4
even with the common factor the sum doesn't = 310 .

Is the question incorrect?
 
  • #5
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:
  • #6
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
 
  • #7
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.
 
  • #8
Thank you...
 
  • #9
i was going down the right track...but must have lost myself with the wrong sign...
thanks :)
 

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

1. What is an arithmetic progression?

An arithmetic progression is a sequence of numbers where each term is obtained by adding a fixed number to the previous term. For example, the sequence 2, 5, 8, 11, 14 is an arithmetic progression with a common difference of 3.

2. How do you find the nth term of an arithmetic progression?

The formula for finding the nth term of an arithmetic progression is:
an = a1 + (n-1)d
where an is the nth term, a1 is the first term, and d is the common difference.

3. Can an arithmetic progression have a negative common difference?

Yes, an arithmetic progression can have a negative common difference. This means that the terms in the sequence will decrease by a fixed amount instead of increasing.

4. What is the sum of an arithmetic progression?

The sum of an arithmetic progression can be calculated using the formula:
Sn = (n/2)(a1 + an)
where Sn is the sum of the first n terms, a1 is the first term, and an is the nth term.

5. How is an arithmetic progression useful in real life?

Arithmetic progressions are useful in many real-life situations, such as calculating interest rates, predicting population growth, and determining the depreciation of assets. They are also used in many mathematical and scientific fields, including physics, chemistry, and economics.

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