If a1,a2,a3,…,a4001are term of an AP such that 1a1a2+1a2a3+…+1a4000a4001=10a2 and a2+a4000=50, then a1−a4001is equal to

Now,1a1a2+1a2a3+…+1a4000a4001=1da2−a1a1a2+a3−a2a2a3+…+a4001−a4000a4000a4001=1d1a1−1a2+1a2−1a3+…+1a4000−1a4001=1d1a1−1a4001=4000a1a4001=10 (given) ⇒ a1a4001=400 ...(i) a1+a4001=a2+a4000=50 .....(ii)∴ a1−a40012=a1+a40012−4a1a4001=(50)2−1600⇒ a1−a4001=30

If a1,a2,a3,…,a4001are term of an AP such that 1a1a2+1a2a3+…+1a4000a4001=10a2 and a2+a4000=50, then a1−a4001is equal to

Now,1a1a2+1a2a3+…+1a4000a4001=1da2−a1a1a2+a3−a2a2a3+…+a4001−a4000a4000a4001=1d1a1−1a2+1a2−1a3+…+1a4000−1a4001=1d1a1−1a4001=4000a1a4001=10 (given) ⇒ a1a4001=400 ...(i) a1+a4001=a2+a4000=50 .....(ii)∴ a1−a40012=a1+a40012−4a1a4001=(50)2−1600⇒ a1−a4001=30

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If $a_{1}, a_{2}, a_{3}, \dots, a_{4001}$ are term of an AP such that $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots + \frac{1}{a_{4000} a_{4001}} = 10 a_{2}$ and $a_{2} + a_{4000} = 50$ , then $|a_{1} - a_{4001}|$ is equal to

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Detailed Solution

Now, $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots + \frac{1}{a_{4000} a_{4001}}$

$\begin{array}{l} = \frac{1}{d} (\frac{a_{2} - a_{1}}{a_{1} a_{2}} + \frac{a_{3} - a_{2}}{a_{2} a_{3}} + \dots + \frac{a_{4001} - a_{4000}}{a_{4000} a_{4001}}) \\ = \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + \dots + \frac{1}{a_{4000}} - \frac{1}{a_{4001}}) \end{array}$

$\begin{array}{l} = \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{4001}}) = \frac{4000}{a_{1} a_{4001}} = 10 (given) \\ \Rightarrow a_{1} a_{4001} = 400 . . . (i) \\ a_{1} + a_{4001} = a_{2} + a_{4000} = 50 . . . . . (ii) \\ ∴ {(a_{1} - a_{4001})}^{2} = {(a_{1} + a_{4001})}^{2} - 4 a_{1} a_{4001} \\ = (50)^{2} - 1600 \\ \Rightarrow |a_{1} - a_{4001}| = 30 \end{array}$