A) -2, 4
B) -4, 2
C) 2, 6
D) none
Correct Answer: B
Solution :
[b] Let the last three numbers in A.P. be a, \[a+6,\] \[a+12,\] then the first term is also \[a+12,\]. But \[a+12,\]a, \[a+6\] are in GP. \[\therefore \,\,\,\,{{a}^{2}}=(a+12)\,(a+6)\Rightarrow {{a}^{2}}={{a}^{2}}+18a+72\] \[\therefore \,\,\,\,a=-4.\] \[\therefore \] The numbers are \[8,\,\,-4,\,\,2,\,\,8.\]You need to login to perform this action.
You will be redirected in
3 sec