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.

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

1 Answer 1


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?

  • $\begingroup$ yes, thank you so much! $\endgroup$
    – borns
    Feb 11, 2020 at 16:16

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