Most Important Selected Qs for JEE AdvancedMathematicsSequences and Series
Let a₁, a₂, a₃ and b₁, b₂, b₃ be arithmetic progressions such that a₁=25, b₁=75 and a₁₀₀+b₁₀₀=100 . Then
Options
- AThe difference between successive terms in progression 'a' is opposite of the difference in progression 'b'.
- Ba_n+b_n=100 for any n .
- C(a₁+b₁ ), (a₂+b₂ ), (a₃+b₃ ) , are in A.P.
- D_ r=1 ¹⁰⁰ (a_r+b_r )=10000
Correct answer
D. _ r=1 ¹⁰⁰ (a_r+b_r )=10000
Step-by-step solution
array ll a₁+ (a₁+d ), & (a₁+2 d ), b₁+ (b₁+d₁ ), & (b₁+2 d₁ ) array hence a₁₀₀=a₁+99 d aligned & b ₁₀₀= b ₁+99 ~d ₁ & add aligned a₁₀₀+b₁₀₀=100+99 (d+d₁ ) hence d+d₁=0 d=-d₁ (A) (B) and (C) are obviously true. _ r=1 ¹⁰⁰ (a_r+b_r )= 100 2 [ (a₁+b₁ )+ (a₁₀₀+b₁₀₀ ) ]= 100 200 2 =10^4 (D) ( using S_n= n 2 (a+d) )