question.

JEE Mains Tutorials › Algebra questions › question.

two arithmetic progressions are a₁, a₁, a₁ ........ and b₁, b₂ , b₃ ......... such that a₁ + b₁ = 100.
also a₂₂ - b₂₁ = b₉₉ - b₁₀₀.
find the sum of 100 terms of the progression
(a₁ + b₁) , (a₂ + b₂) ..........

#Mathematics #Algebra

UP 0 DOWN 0 0 2
Post on FacebookPost anonymously?

2 Answers

Sambit Senapati ·2011-09-05 04:33:12

If it is a₂₂ - a₂₁ = b₉₉- b₁₀₀ (which I am quite sure is the case) the the answer comes out to be 10000

a₂₂ - a₂₁ = b₉₉ - b₁₀₀
=>d₁=-d₂ -(i)

(a₁ + b₁) + (a₂ + b₂) ..........+(a₁₀₀+b₁₀₀)
=(a₁+a₂+...+a₁₀₀) + (b₁+b₂+...+b₁₀₀)
=50(a₁+a₁₀₀)+50(b₁+b₁₀₀)
=50(a₁+b₁+a₁₀₀+b₁₀₀)
=50(a₁+b₁+a₁+99d₁+b₁+99d₂)
=50{2(a₁+b1)+99(d₁+d₂)}
=50{2*100+99*0} (From (i))
=50*200=10000