a1+a2=4⇒a1+a1r=4…(1)
a3+a4=16⇒a1r2+a1r3=16…(2)
r21=41⇒r2=4⇒r=−2(a1<0)
i=1∑aai=r−1a1(r9−1)=(−2−1)(−4)[(−2)9−1]=34(−513)=4λ
λ=−171.
JEE Main 2020 — Mathematics Algebra
Let a1,a2,a3,…, be a G.P. such that a1<0,a1+a2=4 and a3+a4=16. If i=1∑9ai=4λ, then λ, is equal to.
Held on 7 Jan 2020 · Verified 6 Jul 2026.
−513
−171
171
3511
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