Question

( left|begin{array}{lll}a+1 & a+2 & a+p a+2 & a+3 & a+q a+3 & a+4 & a+gammaend{array}right|=0 )
( C_{2} rightarrow C_{2}-C_{1} )
( left|begin{array}{ccc}a+1 & 1 & a+p a+2 & 1 & a+a a+3 & 1 & a+rend{array}right|=0 )
Only possible when ( P, q ), ( y ) are in
( P= )
( A )
because tura ption
colemn will become same then
( Rightarrow d e t=0 )

# a+1 a+2a+pl TC a +2 a +3 a+9 a+3 a+4 atr (a) AP If = 0, then p, q, r are in (b) GP (d) None of these (c) HP

Solution