Question

Let $<a_n>$ be an arithmetic sequence such that arithmetic mean of $a_1,a_3,a_5,\dots \dots ,a_{97},a_{99}$ is 1 Find the value of $\left|\sum _{r=1}^{50}(-1{)}^{\frac{r(r+1)}{2}}\cdot a_{2r-1}\right|$.

Answer 1

Answer: 2

$a_1+a_3+\dots +a_{99}=50\Rightarrow a+a+2d+a+4d+\dots a+98d=50$
$50a+2d(1+2+\dots +49)=50\Rightarrow 50a+\frac{2d(50)(49)}{2}=50$
$a+49d=1$
$a_{50}=1$
$\left|\sum _{r=1}^{50}(-1{)}^{\frac{r(r+1)}{2}}\cdot a_{2r-1}\right|$
$-a_1-a_3+a_5+a_7-a_9-a_{11}+\dots +a_{93}+a_{95}-a_{97}-a_{99}$
$\text{ (26 negative terms and }24\text{ positive) }$
$\Rightarrow \left|-a_1-a_{99}\right|=|-2a-98d|\Rightarrow |-2|\Rightarrow 2$