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 .