Summation of 3n + 2 from n = 1 to n = 14 = (3(1) + 2) + (3(2) + 2) + (3(3) + 2) + . . . + (3(14) + 2) = 5 + 8 + 11 + ... + 44 ia an arithmetic progression with first term (a) = 5, common difference (d) = 3 and last term (l) = 44 and n = 14
Sn = n/2(a + l) = 14/2(5 + 44) = 7(49) = 343
Therefore, the required summation is 343.
X=2 because when you multiply 8 x 2=16 - 4=12
8(2)-4=12
A) use a straightedge to draw a line thought D and F
Study all notes, reread the chapters again. Have someone ask questions on the chapters page by page. This always has worked for me. Plus try to do this again the night before the test. You will be surprised how much you can remember by doing it again the night before the test. Hope this helps.
