# Arithmetic Sequence Equation

Find a solution for: $$\left( x+1 \right) + \left( x+4 \right) + \left( x+7 \right) +...+ \left( x+28 \right) = 155$$

I tried with $$S_n$$ formula, $$S_n = 155 = \frac{n[(x+1)+(3n-2)]}{2}$$ But I suppose it's not right.

I tried also putting numbers in $$x$$ from $$1$$ and so on until the correct answer, and I got an answer, but I want to know an easier way when that random number will be higher than a digit.

• In this case, you can determine from the given numbers what $n$ is...! Oct 15 at 15:28
• There should be $3n-\color{red}{3}$ in there (probably a typo). Also, as Greg said, you can find $n$ from the given data by the formula $$a_n=a+(n-1)d$$ Oct 15 at 15:29
• You've lost $x$ in $x+3n-2$ in your formula. Oct 15 at 15:31

The sum can be written as $$\sum_{k=0}^9(x+1+3k)$$ Now the sum of consecutive terms of an arithmetic sequence is the average of the first and the last term, times the number of terms. Namely, in the present case: $$\frac{(x+1)+(x+28)}2\times 10=(2x+29)\times 5=10x+145,$$ so we obtain $$10x+145=155\iff 10x=10\iff x=1.$$
• In this solution, you already know $n$ which isn't a case in the problem. Oct 16 at 17:28
• There is a much easier way, that I have found now. $$Sn = 155 = \frac{n[(x+1) + (x+28)]}{2}$$ $$310 = n(2x+29)$$ From that it's easy to guess $x$ and $n$ but IDK if that's correct way to solve,even the answer is right. Oct 16 at 17:33
$$(x+1)+(x+4)+(x+7)+(x+10)+(x+13)+(x+16)+(x+19)+(x+22)+(x+25)+(x+28)=155$$ $$\implies$$ $$10x+145=155$$ $$\implies$$ $$10x=10$$ $$\implies$$ $$x=1$$