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

# If ${\mathrm{a}}_{1},{\mathrm{a}}_{2},{\mathrm{a}}_{3},\dots ,{\mathrm{a}}_{4001}$are term of an AP such  that $\frac{1}{{\mathrm{a}}_{1}{\mathrm{a}}_{2}}+\frac{1}{{\mathrm{a}}_{2}{\mathrm{a}}_{3}}+\dots +\frac{1}{{\mathrm{a}}_{4000}{\mathrm{a}}_{4001}}=10{\mathrm{a}}_{2}$ and , then $\left|{\mathrm{a}}_{1}-{\mathrm{a}}_{4001}\right|$is equal to

1. A

20

2. B

30

3. C

40

4. D

None of these

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### Solution:

Now,$\frac{1}{{\mathrm{a}}_{1}{\mathrm{a}}_{2}}+\frac{1}{{\mathrm{a}}_{2}{\mathrm{a}}_{3}}+\dots +\frac{1}{{\mathrm{a}}_{4000}{\mathrm{a}}_{4001}}$

$\begin{array}{l}=\frac{1}{\mathrm{d}}\left(\frac{{\mathrm{a}}_{2}-{\mathrm{a}}_{1}}{{\mathrm{a}}_{1}{\mathrm{a}}_{2}}+\frac{{\mathrm{a}}_{3}-{\mathrm{a}}_{2}}{{\mathrm{a}}_{2}{\mathrm{a}}_{3}}+\dots +\frac{{\mathrm{a}}_{4001}-{\mathrm{a}}_{4000}}{{\mathrm{a}}_{4000}{\mathrm{a}}_{4001}}\right)\\ =\frac{1}{\mathrm{d}}\left(\frac{1}{{\mathrm{a}}_{1}}-\frac{1}{{\mathrm{a}}_{2}}+\frac{1}{{\mathrm{a}}_{2}}-\frac{1}{{\mathrm{a}}_{3}}+\dots +\frac{1}{{\mathrm{a}}_{4000}}-\frac{1}{{\mathrm{a}}_{4001}}\right)\end{array}$

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