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.
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 $2(2n)^2+(2n)=\color{red}8n^2+2n$.
Can you solve it now?