Let a1, a2 , a3 , ... be in an A.P. such that summation k=1 to 12 a(2k-1) =-72/5 a1, a1 is jee main

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Q) Let 𝑎_1,𝑎_2,𝑎_3…. be in an A.P. such that ∑_(𝑘=1)^12 𝑎_(2𝑘−1)=−72/5 𝑎_1,𝑎_1≠0. If ∑_(𝑘=1)^𝑛 𝑎_

Let a1, a2, a3... be in an AP. such that ∑ a2k-1=-72/5a1, a1≠0. If ∑ak=0, then n is:

Let a1,a2,a3...be in an A.P. such that Σ a2k-1=-72/5 a1,a≠0.If Σak=0,then n is :

Let a1,a2,a3… in an AP and Sum_k=1 to 12 a_(2k-1) =(-72/5)a1 and Sum_k=1 to n a_k =0 |jeemain2025

Let a_1,a_2,a_3,…be in an A.P. such that ∑_(k=1)^12▒a_(2k-1) =-72/5 a_1,a_1≠0. If ∑_(k=1)^n▒a_k=0,

𝑎1, 𝑎2,… A.P.such that ∑_(k=1)^12▒a_(2k-1) = 1 72/5  a_1,a_1 ≠ 0, if ∑_(k=1)^n▒a_k =[JEE(M) 2/03/25

Let a1, a2, a3,.. be in an A.P. such that121a2k-1=- a1, a1 # 0. IfLk=1 ak = O, then n is: || Let's

Arithmetic Progression: Let a1, a2, a3, … be an A.P. If(𝒂_𝟏+𝒂_𝟐+…𝒂_𝟏𝟎)/(𝒂_𝟏+𝒂_𝟐+…+𝒂_𝒑 ) = 𝟏𝟎𝟎/𝒑^𝟐