If `a_(1),a_(2),a_(3),….,a_(r)` are in GP, then prove that the determinant `|(a_(r+1),a_(r+5),a_(r+9)),(a_(r+7),a_(r+11),a_(4+15)),(a_(r+11),a_(r+17),a_(r+21))|` is independent of `r`.

Answer

N/a

Explanation

We know that `a_(r+1)=AR^((r+1)-1)=AR^(r)`
where `r=r` th term of GP `A=` First term of a GP and `R=` Common ratio of GP
We have `|(a_(r+1),a_(r+5),a_(r+9)),(a_(r+7),a_(r+11),a_(r+15)),(a_(r+11),a_(r+17),a_(r+21))|`
`=|(AR^(r),AR^(r+4),AR^(r+8)),(AR^(r+6),AR^(r+10),AR^(r+14)),(AR^(4+10),AR^(r+16),AR^(r+20))|`
`=AR^(r).AR^(r+6).AR^(r+10)|(1,AR^(4),AR^(8)),(1,AR^(4),AR^(8)),(1,AR^(6),AR^(10))|`
[takng `AR^(r),AR^(r+6)` and `AR^(r+10)` common from `R_(1),R_(2)` and `R_(3)` respectivley]
`=0` [since `R_(1)` and `R_(2)` are identicals]