First slide

Arithmetic progression

Question

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

Moderate

A

20
B

30
C

40
D

None of these

Solution

$\begin{array}{l} now, \frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots + \frac{1}{a_{4000} a_{4001}} \\ = \frac{1}{d} (\frac{a_{2} - a_{1}}{a_{1} a_{2}} + \frac{a_{3} - a_{2}}{a_{2} a_{3}} + \dots + \frac{a_{4001} - a_{4000}}{a_{4000} a_{4001}}) \\ = \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + \dots + \frac{1}{a_{4000}} - \frac{1}{a_{4001}}) \end{array}$
$= \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{4001}}) = \frac{4000}{a_{1} a_{4001}} = 10$ Given
$\begin{matrix} \Rightarrow a_{1} a_{4001} = 400 - - - - (i) \\ ∴ a_{1} + a_{4001} = a_{2} + a_{4000} = 50 - - - (ii) \\ ∴ {(a_{1} - a_{4001})}^{2} = {(a_{1} + a_{4001})}^{2} - 4 a_{1} a_{4001} \\ = (50)^{2} - 1600 \end{matrix}$
$\Rightarrow |a_{1} - a_{4001}| = 30$

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