The solution of equation $4+6+8+10+\cdots +x =270$ is 15. The solution of equation 
$4+6+8+10+\cdots +x =270$ 
is $x=15$. 
How can I prove it? 
I ve tried the geometric sequence but I cannot figure out the pattern.
 A: Dividing both sides by 2 gives you:
$$2 + 3 + 4 + 5 ... + x/2 = 135$$
Adding 1 to both side of the equation then gives you:
$$1 + 2 + 3 + 4 ... + x/2 =136$$
Since the sum of all integers from 1 to $n$ is defined as:
$$\frac{n(n+1)}{2}$$
We can rewrite the equation as:
$$\frac{x(x+2)}{8}= 136$$
$$x^2 + 2x - 1088 = 0$$
$$(x+34)(x-32)=0$$
Since $x$ must be positive:
$$x = 32$$
A: The solution is incorrect. In order for the question to make sense, we need to assume that $x=2n, n \in \mathbb N$ and $x \geq 4$.
In that case what you have is a sum of an arithmetic series.
The equation you get is
$$270 = \frac{x+4}{2}(1+\frac{x-4}{2})$$
This is a quadratic equation, solving it yields two solutions $x=32$ and $x=-34$ but of course you are only interested in the first one.
A: $$4+6+8+10+\cdots +x=270\\ \frac { 4+x }{ 2 } \left( \frac { x-4 }{ 2 } +1 \right) =270\\ \frac { \left( 4+x \right)  }{ 2 } \frac { \left( x-2 \right)  }{ 2 } =270\\ \left( x-2 \right) \left( x+4 \right) =1080\\ { x }^{ 2 }+2x-1088=0\\ x=-1\pm \sqrt { 1089 } \\ x=32$$
A: Since $$1+2+3+4+.....+(n-1)+n = \frac{n(n+1)}{2}$$
$$2+4+6+8+.....+2(n-1)+2n = n(n+1)$$
That is $$4+6+8+.....+2(n-1)+2n = n(n+1)-2$$
Thus $$n(n+1)-2 \leq 270$$
$$n^2+n-272 \leq 0$$
$$(n+17)(n-16) \leq 0$$
So $$n=16$$
Thus $$4+6+8+10+12+14+16+18+20+22+24+26+28+30+32=270$$
I think the question should be $$(4+6+8+...)\cdot x =270$$
Then we can argue that factors of $270$ are $1,2,3,5,6,9,10,15,18,27,30,45,54,90,135,270$
Then the only chances are 
$$4+6 =10 $$ which gives $x=27$
$$4+6+8 = 18$$ which gives $x=15$
$$4+6+8+10+12+14 =54 $$ which gives $x=5$
$$4+6+8+10+12+14+16+18+20+22+24+26+28+30+32 = 270 $$ which gives $x=1$
A: $x=15$ is not a solution to this equation. The question does NOT ask how many terms are in the series although it is the answer to that question that is given.
Furthermore, there is nothing in the question stating that $x$ is a term of the arithmetic series. If $x$ is a term of the series then it must be the 15th term. But as stated $x$ could be 0 since 
\begin{equation}
4+6+8+10+ \cdots + 30 + 32 + 0=270
\end{equation}
But $x=32$ is also a solution since
\begin{equation}
4+6+8+10+ \cdots + 30 + 32 =270
\end{equation}
and $x=62$ is a solution since 
\begin{equation}
4+6+8+10+ \cdots + 28 + 62 =270
\end{equation}
etc.
The solution set for $x$ is $\{270-n(n+3) \,\vert\, 4\le n\le 15\}$.
