Help please? The first 9 terms of an arithmetic series have a sum of 99. The common difference is 2.What are the first three terms of the series? Show all work.

Question

cwrw238 · Answer

shoot! what happened to my reply!!!
first 3 terms are 3 5 and 7

cwrw238 · Answer

sum 9 terms = 9/2[2a + 8*2] = 99
solving for a gives
a=first term = 3

anonymous · Answer

There is probably an easier way to do this, but I'm sort of inclined toward the brute force approach to problems of this kind.$$\sum_{k=1}^{9}a_k=x+x+2+x+4+x+6+x+8+x+10+x+12+x+14+x+16$$$$=9x+72=99\implies 9x=27\implies x=3$$So the first three terms are 3,5,7.

anonymous · Answer

why would it be 8*2 cwrw? Can u break it down to how u got the whole answer please?

cwrw238 · Answer

shoot - this website - a pain isn't it?

again:!!!
sum on n terms is given by 

Sn = (n/2)[2a + (n-1)d]

here Sn = 99, n=9 and d = 2  so:

99 = (9/2)[2a + (9-1)2]