Originally Posted by
Iainwb.13 Question -
The sum of the first three terms of an arithmetic sequence is 12. The sum of their squares is 66. Find the fourth term
I know just by trial an error that the first three terms are 1, 4 and 7. therefore the 4th term is 10.
Although I am able to figure this out, there IS an algebraic way of determining the values to prove that these ar , in fact, the numbers of the sequence.
I have started along the lines of
a + (a+d)^2 + (a+2d)^2 = 66
a + a +(a+d) + (a+d) + (a+2d) + (a+2d) = 66
6a+6d = 66
but im not sure where to go from here. i know im on the right track i just cant figure this out. please help !?