If $$a1$$,$$a2$$,$$a3$$,….$$a4001$$ are terms of an AP such that $$1a1a2+1a2a3+$$…$$+1a4000a4001=10$$ and $$a1+a4001=50$$ then $$a1$$−$$a4001$$ is equal to

$$1da4001$$−$$a1a1a4001=10$$⇒$$4000a1a4001=10$$ ∵$$a4001=a1+4000d$$⇒$$1da4001-a1=4000$$⇒$$a1a4001=400Also$$ given $$a1+a4001=50$$∴$$a1$$−$$a40012=a1+a40012$$−$$4a1a4001$$⇒$$a1$$−$$a4001=30$$

Arithmetic progression

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

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Solution

$\begin{array}{l} \frac{1}{d} [\frac{a_{4001} - a_{1}}{a_{1} a_{4001}}] = 10 \\ \Rightarrow \frac{4000}{a_{1} a_{4001}} = 10 (∵ a_{4001} = a_{1} + 4000 d \Rightarrow \frac{1}{d} (a_{4001} - a_{1}) = 4000) \\ \Rightarrow a_{1} a_{4001} = 400 \\ Also given a_{1} + a_{4001} = 50 \\ ∴ {(a_{1} - a_{4001})}^{2} = {(a_{1} + a_{4001})}^{2} - 4 a_{1} a_{4001} \\ \Rightarrow |a_{1} - a_{4001}| = 30 \end{array}$

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Arithmetic progression

Remember concepts with our Masterclasses.

If a1,a2,a3,….a4001 are terms of an AP such that 1a1a2+1a2a3+…+1a4000a4001=10 and a1+a4001=50 then a1−a4001 is equal to

1da4001−a1a1a4001=10⇒4000a1a4001=10 ∵a4001=a1+4000d⇒1da4001-a1=4000⇒a1a4001=400Also given a1+a4001=50∴a1−a40012=a1+a40012−4a1a4001⇒a1−a4001=30

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