Let a 1 , a 2 , a 3 , … , be a G . P . such that a 1 < 0 , a 1 + a 2 = 4 and a 3 + a 4 = 16 . If…

Let a1,a2,a3,, be a G.P. such that a1<0,a1+a2=4 and a3+a4=16. If i=19ai=4λ, then λ, is equal to.
  1. -513
  2. -171
  3. 171
  4. 5113

Solution

a1+a2=4a1+a1r=41

a3+a4=16a1r2+a1r3=162

1r2=14r2=4 r=-2  a1<0

i=1aai=a1r9-1r-1=-4-29-1-2-1=43-513=4λ

λ=-171.

Asked in: JEE Main 2020 (07 Jan Shift 2)

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