The question is this: in the arithmetic sequence $$3,7,11,...$$, find the least value of $$n$$ for which the sum of the first $$2n$$ terms will exceed the sum of the first $$n$$ terms by $$155$$.

What is wrong with my solution?

Sum of $$n$$ terms: $$n/2( 4n+2)$$

Sum of $$2n$$ terms: $$4n^2+2n$$

Final equation: $$2n^2+n-155=0$$

I am not getting $$5$$ for $$n$$; please help.

• the formula for the sum of $n$ terms should depend on the initial term ($3$ in this case) Feb 11, 2020 at 15:56
• please solve it, ive been struggling for a while now Feb 11, 2020 at 15:57
• when $n=1$, what is the sum of the first $2n$ terms? Feb 11, 2020 at 16:07

What is wrong with your solution is that the sum of the first $$n$$ terms of the arithmetic progression
$$3,7,11,...$$ with initial term $$3$$ and difference $$4$$ is $$\dfrac n2(2\times3+(n-1)4)=2n^2+n,$$
so the sum of the first $$2n$$ terms is $$2(2n)^2+(2n)=\color{red}8n^2+2n$$.