A) 4, 20, 36
B) 4, 12, 36
C) 4, 20, 100
D) None of the above
Correct Answer: C
Solution :
\[a,\ ar,\ a{{r}^{2}}\] are in G.P. \[a,\ ar-8,\ a{{r}^{2}}-64\] are in A.P., we get \[\Rightarrow \]\[a({{r}^{2}}-2r+1)=64\] .....(i) Again, \[a,\ ar-8,\ a{{r}^{2}}-64\] are in G.P. \[\therefore \]\[{{(ar-8)}^{2}}=a(a{{r}^{2}}-64)\] or \[a(16r-64)=64\] .....(ii) Solving (i) and (ii), we get\[r=5,\ a=4\]. Thus required numbers are 4, 20, 100. Trick: Check by alternates according to conditions (a) \[\Rightarrow \]4, 20, - 28 which are not in A.P. (b) \[\Rightarrow \]4, 12, - 28 which are also not in A.P. (c) \[\Rightarrow \]4, 20, 36 which are obviously in A.P. with 16 as common difference. These numbers also satisfy the second condition \[i.e.\] 4, 20 - 8, 36 are in G.P.
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