Question

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

Answer 1

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