Question

If a₁, a₂, a₃,..., a₄₀₀₁ are terms of an AP such that $\frac{1}{a_{1}a_{2}}$+$\frac{1}{a_{2}a_{3}}$+….+$\frac{1}{a_{4000}a_{4001}}$= 10 and a₂ + a₄₀₀₀ = 50, then |a₁ – a₄₀₀₁| is equal to –

• A. 20
• B. 30
• C. 40
• D. None of these

Answer 1

Answer: (B)

30

Answer 2

Now, $\frac{1}{a_{1}a_{2}} + \frac{1}{a_{2}a_{3}}$+….+$\frac{1}{a_{4000}a_{4001}}$ = $\frac { 1 } { \mathrm {~d} }$

$\left( \frac{a_{2} - a_{1}}{a_{1}a_{2}} + \frac{a_{3} - a_{2}}{a_{2}a_{3}} + .... + \frac{a_{4001} - a_{4000}}{a_{4000}a_{40001}} \right)$

= $\left( \frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + .... + \frac{1}{a_{4000}} - \frac{1}{a_{4001}} \right)$

= $\left( \frac{1}{a_{1}} - \frac{1}{a_{4001}} \right)$=$\frac{4000}{a_{1}a_{4001}}$= 10 (given)

̃ a₁ a₄₀₀₁ = 400 …(i)

a₁ + a₄₀₀₁ = a₂ + a₄₀₀₀ = 50 …(ii)

\ (a₁ – a₄₀₀₁)² = (a₁ + a₄₀₀₁)² – 4a₁a₄₀₀₁

= (50)² – 1600

̃ |a₁ –a₄₀₀₁| = 30.