Mark the correct alternative in each of the following:

The first three of four given numbers are in G.P. and their last three are in A.P. with common difference 6. If first and fourth numbers are equal, then the first number is


Let, the last three numbers of the set which are in A.P be b,b+6,b+12 and the first number be a.


Thus, the four numbers are a,b,b+6,b+12


Given:


a=b+12…(1)


Also, given a,b,b+6 are in G.P


From equation (1)


b+12,b,b+6 are in G.P



b2= (b+6)(b+12)


b2= b2+18b+72


18b= -72


b= -4


a= -4+12


=8


Hence the four numbers are 8,-4,2,8

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