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Question

If the lines ax+ky+10=0,bx+(k+1)y+10=0 and cx+(k+2)y+10=0 are concurrent, then:


A

a,b,c are in G.P.

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B

a,b,c are in H.P.

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C

a,b,c are in A.P.

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D

(a+1)2=c

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Solution

The correct option is C

a,b,c are in A.P.


Explanation for correct option

Given lines, ax+ky+10=0,bx+(k+1)y+10=0 and cx+(k+2)y+10=0 are concurrent.

So, the determinant will be zero.

ak10bk+110ck+210=0

Applying the row transformation operations,

R2R2-R1R3R3-R1

ak10b-a10c-a20=0

Upon solving, we get,

a(0-0)-k(0-0)+10(2b-2a-c+a)=02b-a-c=0a+c=2b

So, the condition obtained is of Arithmetic progression.

Therefore, option (C) is the correct answer.


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