Let a_1,a_2,,a_n be a given A.P.

Question

Let $a_1,a_2,\ldots,a_n$ be a given A.P. whose common difference is an integer, and $S_n=a_1+a_2+\cdots+a_n$. If $a_1=1$, $a_n=300$, and $15\le n\le50$, then the ordered pair $(S_{n-4}, a_{n-4})$ is equal to:

Accepted Answer

$300 = 1+(n-1)d \Rightarrow d = \dfrac{299}{n-1} = \dfrac{13	imes23}{n-1}$, requiring $n-1\in\{13,23\}$ (integer $d$), so $n=14$ or $24$; within $[15,50]$, $n=24$, $d=13$.

$n-4=20$: $a_{20} = 1+19(13) = 248$, $S_{20} = \dfrac{20}{2}(2+19	imes13) = 2490$.

$(S_{20}, a_{20}) = (2490, 248).$

Solutions · 1