Find Number of Terms

Nov 2019
23
2
NYC
This question comes from chapter 9 in terms of arithmetic sequences.

MathMagic191112_1.png
 

romsek

MHF Helper
Nov 2013
6,671
3,005
California
$a_k = 8+3k,~k=1,2,\dots$

$\sum \limits_{k=1}^n~a_k = \sum \limits_{k=1}^n~(8+3k) = \\

8n + 3 \dfrac{n(n+1)}{2} = \\

8n + \dfrac 3 2 n^2 + \dfrac 3 2 n = \\

\dfrac 3 2 n^2 +\dfrac{19}{2}n = 1092
$

$3n^2 + 19n - 2184 = 0$

$n = \dfrac{-19 \pm \sqrt{(19)^2 + 4(3)(2184)}}{6} = \left(-\dfrac{91}{3},~24\right)$

Obviously we ignore the negative solution to obtain

$n = 24$