Question

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

Answer 1

The correct option is C 4, 20, 100

$a,ar,a{r}^2$ 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 $a(16r-64)=64$ ....(i)
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.
(b) $\Rightarrow$ 4,20, 36 which are obviosly 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.