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Let a(1),a(2),a(3), …, a(10) be in G.P. ...

Let `a_(1),a_(2),a_(3), …, a_(10)` be in G.P. with `a_(i) gt 0` for i=1, 2, …, 10 and S be te set of pairs (r, k), r, k `in` N (the set of natural numbers)
for which `|(log_(e)a_(1)^(r)a_(2)^(k),log_(e)a_(2)^(r)a_(3)^(k),log_(e)a_(3)^(r)a_(4)^(k)),(log_(e)a_(4)^(r)a_(5)^(k),log_(e)a_(5)^(r)a_(6)^(k),log_(e)a_(6)^(r)a_(7)^(k)),(log_(e)a_(7)^(r)a_(8)^(k),log_(e)a_(8)^(r)a_(9)^(k),log_(e)a_(9)^(r)a_(10)^(k))|` = 0. Then the number of elements in S is

A

10

B

Infinitely many

C

2

D

4

Text Solution

Verified by Experts

The correct Answer is:
B
All the term of determinant are of the form
In `a_(n)^(r) a_(n+1)^(k)`
Since `a_(1), a_(2), a_(3), ....G.P.`
Let `therefore a_(n)=AR^(n-1)`
In `(a_(n))^(r) (a_(n+1))^(k)= 4"In"a_(n)=k log_(e)a_(n+1)=r"In" AR^(n-1) +k"In"AR^(n)`
`= (r+k)"In"A+((n-1)r+nk)"In"R`
Then the det given in question will become
`|((r+k)"In"A+k"In"R,(r+k)"In"A+(r+2k)"In"R,(r+k)"In"A+(2r+3k)"In"R),((r+k)"In"A+(3r+4k)"In"R,(r+k)"In"A+(4r+5k)"In"R,(r+k)"In"A+(5r+6k)"In"R),((r+k)"In"A+(6r+7k)"In"R,(r+k)"In"A+(7r+8k)"In"R,(r+k)"In"A+(8r+9k)"In"R)|`
Now `R_(2) to R_(2) -R_(1)`
`" " R_(3) to R_(3) -R_(1)`
`|((r+k)"In"A+k"In"R,(r+k)"In"A+(r+2k)"In"R,(r+k)"In"A+(2r+3k)"In"R),(3(r+k)"In"R,3(r+k)"In"R,3("In"R)(r+k)), (6(r+k)"In"R, 6(r+k)"In"R, (6"In"R)(r+k))|`
Now Row 3 and Row 2 are proportion
`therefore ` This determinant will always be zero irrespective of values of r & k
`therefore ` No. of pairs of (r, k) will be infinite
Hence 2 is the correction option
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