Question

Find the least value of n such that 2+5+8+11+...to n terms >=200

Answer 1

$2+5+8+11+\dots +(3n-1)\geq200\\\\\dfrac {n}{2}[2+3n-1]\geq200\\\\\dfrac {n}{2}[3n+1]\geq200\\\\3n^2+n\geq400\\\\3n^2+n-400\geq0\\\\\implies\ n\in\left (-\infty,\ \dfrac {-1-\sqrt{1^2-(4\cdot3\cdot(-400))}}{2\cdot3}\right]\cup\left [\dfrac {-1+\sqrt{1^2-(4\cdot3\cdot(-400))}}{2\cdot3},\ \infty\right)\\\\n\in\left (-\infty,\ \dfrac {-1-\sqrt{4801}}{6}\right]\cup\left [\dfrac {-1+\sqrt{4801}}{6},\ \infty\right)$

Taking $\sqrt{4801}\approx69,$

$n\in\left (-\infty,\ \dfrac {-1-69}{6}\right]\cup\left [\dfrac {-1+69}{6},\ \infty\right)\\\\n\in\left (-\infty,\ \dfrac {-35}{3}\right]\cup\left [\dfrac {34}{3},\ \infty\right)$

But we know that $n>0.$

$\therefore\ n\in\left [\dfrac {34}{3},\ \infty\right)\\\\n\in\left [11.\overline{3},\ \infty\right)$

So we can say that the least possible value of $n$ is $\large\textbf{12}.$