mark-ashleigh
Apr30-05, 11:58 AM
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 .
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 .