Let $\quad \mathrm{a}_{\mathrm{b}, 2} \mathrm{a}_2, \mathrm{a}_3, \ldots$ be an A.P of non-zero numb

Question

Let $\quad \mathrm{a}_{\mathrm{b}, 2} \mathrm{a}_2, \mathrm{a}_3, \ldots$ be an A.P of non-zero numbers If $\frac{1}{a_1 a_2}+\frac{1}{a_2 a_3}+\ldots+\frac{1}{a_{100} a_{101}}=\frac{1}{6}$ and $a_1+a_{101}=50$ then $\left|a_1-a_{101}\right|$ is equal to

Professor PC Thomas Classes · Answer

Given \ $ \begin{aligned} & =\frac{1}{d}\left[\left(\frac{1}{a_1}-\frac{1}{a_2}ight)+\left(\frac{1}{a_2}-\frac{1}{a_3}ight)+\ldots+\left(\frac{1}{a_{100}}-\frac{1}{a_{101}}ight)ight]=\frac{1}{6} \ & =\frac{1}{d}\left[\frac{1}{a_1}-\frac{1}{a_{101}}ight]=\frac{1}{6} \ & =\frac{1}{d}\left[\frac{a_1+100 d-a_1}{a_1 a_{101}}ight]=\frac{1}{6}, \ & \frac{1}{a_1 a_{101}}=\frac{1}{600} \ & a_1 a_{101}=600, a_1+a_{101}=50 \end{aligned} \ $ $ \  \begin{aligned} & \left(a_1-a_{101}ight)^2=\left(a_1+a_{101}ight)^2-4 a_1 a_{101} \ & =2500-2400=100 \end{aligned} $