• # question_answer Three numbers form a G.P. If the ${{3}^{rd}}$ term is decreased by 64, then the three numbers thus obtained will constitute an A.P. If the second term of this A.P. is decreased by 8, a G.P. will be formed again, then the numbers will be A) 4, 20, 36 B) 4, 12, 36 C) 4, 20, 100 D) None of the above

$a,\ ar,\ a{{r}^{2}}$ 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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