Arithmetic progression Trouble

In summary, the conversation discusses finding the possible values of x in an arithmetic progression, with the given information of the first two terms and the sum of the first ten terms being 310. The participant mentions using binomial expansion and the quadratic theorem to find values for x, and then finding the difference between the first two terms. They also mention calculating the tenth term and using the formula for the sum of an arithmetic progression. They end by asking for clarification on their calculations.
  • #1
mark-ashleigh
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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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  • #3


It seems like you are on the right track with your approach to solving the problem. However, it's possible that you made a mistake in your calculations somewhere along the way. It's always a good idea to double check your work and make sure all of your calculations are accurate.

As for the value of 236.7, it's possible that you may have used the wrong formula for the sum. The formula you used, S10 = 10 x (a + l)/2, is the formula for the sum of an arithmetic series with an odd number of terms. Since we are dealing with 10 terms, which is an even number, we need to use a different formula: S10 = 10 x (2a + (n-1)d)/2.

Using this formula with the values you calculated for the first term, difference, and tenth term, we should get a sum of 310. So it's possible that the mistake lies in using the wrong formula.

In any case, it's important to carefully check your work and make sure all of your calculations are accurate. If you are still having trouble, don't hesitate to ask for help from your teacher or a tutor. With practice and patience, you will be able to master arithmetic progression and succeed on your exam. Good luck!
 

FAQ: Arithmetic progression Trouble

What is an arithmetic progression?

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

What is the formula for finding the nth term in an arithmetic progression?

The formula for finding the nth term in 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.

How do you determine if a sequence is an arithmetic progression?

To determine if a sequence is an arithmetic progression, you can check if the difference between consecutive terms is the same. If the difference is constant, then the sequence is an arithmetic progression.

What is the sum of an arithmetic progression?

The sum of an arithmetic progression can be found 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.

How can I use arithmetic progressions in real life?

Arithmetic progressions can be used in various real-life situations, such as calculating the amount of money saved over time with a fixed monthly deposit, determining the number of steps taken in a regularly increasing exercise routine, or predicting population growth over a period of time.

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