Let a1, a2, a3, ……. a101 are in A.P. such that \frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \ldots \ldots . + \frac{1}{a_{100} a_{101}} = 10 & a2 + a100 = 50, then the value of (a1

Engineering

Mathematics

Arithmetic Progression and Its Properties

Question

Let a₁, a₂, a₃, ……. a₁₀₁ are in A.P. such that $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \dots \dots . + \frac{1}{a_{100} a_{101}}$ = 10 & a₂ + a₁₀₀ = 50, then the value of (a₁ – a₁₀₁)² is equal to

2500

4950

5050

2460

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\frac{1}{d} (\frac{a_{2} - a_{1}}{a_{1} a_{2}} + \frac{a_{3} - a_{2}}{a_{2} a_{3}} + \dots \dots . + \frac{a_{101} - a_{100}}{a_{100} a_{101}}) = 10$
$∴ \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}} + \dots \dots . + \frac{1}{a_{100}} - \frac{1}{a_{101}}) = 10$
$∴ \frac{1}{d} (\frac{1}{a_{1}} - \frac{1}{a_{101}}) = 10 \Rightarrow \frac{a_{101} - a_{2}}{a_{1} \cdot a_{101}} = 10 d$
⇒ a₁ · a₁₀₁ = 10. Also a₁ + a₁₀₁ = 50
$∴$ (a₁ – a₁₀₁)² = (a₁ + a₁₀₁)² – 4a₁a₁₀₁ = 2500 – 40 = 2460. Ans.