Question

If the sum of three numbers in G.P. is 56 . If we subtract $1,7,21$ from these numbers in that order, we obtain an Arithmetic Progression (A.P.), then which of the following are true?
I. The numbers are $8,16,32$.
II. The numbers are $32,16,8$.
III. The numbers are $-8,16,-32$.
IV. The numbers are $-32,16,-8$.

• A. I and II are true
• B. III and IV are true
• C. I and III are true
• D. II and IV are true

Answer 1

Answer: (A)

Let three numbers in G.P. are $a, a r, a r^2$.
Given, $ a+a r+a r^2=56 ....$(i)
Again, $a-1, a r-7, a r^2-21$ are in A.P.
$\Rightarrow 2(a r-7) =(a-1)+\left(a r^2-21\right) $
$ (\because \text { if } a, b, c \text { are in } \wedge . P ., \text { then } 2 b-a+c)$
$\Rightarrow 2 a r-14 =a+a r^2-22$
$\Rightarrow a+a r^2-2 a r =-14+22 $
$\Rightarrow a+a r^2-2 a r =8 ....$(ii)
Dividing Eq. (i) by Eq. (ii), we get
$ \frac{a+a r+a r^2}{a+a r^2-2 a r}=\frac{56}{8} \Rightarrow \frac{1+r+r^2}{1+r^2-2 r}=\frac{7}{1}$
$ \Rightarrow 1+r+r^2=7+7 r^2-14 r $
$ \Rightarrow 6 r^2-15 r+6=0$
Dividing by 3 , we get
$\Rightarrow 2 r^2-5 r+2=0$
Factorizing it by splitting the middle term, we get
$ 2 r^2-(4+1) r+2 =0$
$\Rightarrow 2 r^2-4 r-r+2 =0$
$\Rightarrow 2 r(r-2)-(r-2) =0$
$\Rightarrow (r-2)(2 r-1) =0$
$\Rightarrow r =2, \frac{1}{2}$
If $r=2$, then from Eq. (i), we get
$a+2 a+4 a =56$
$\Rightarrow 7 a =56$
$\therefore a =8$
Then, numbers are
$ a =8$
$\Rightarrow a r =8 \times 2=16 $
$\Rightarrow a r^2 =8 \times 4=32$
Hence, required numbers are $8,16,32$.
If $r=\frac{1}{2}$, then from Eq. (i), we get
$ \frac{a}{1}+\frac{a}{2}+\frac{a}{4}=56$
$ \Rightarrow \frac{4 a+7 a+1}{4}=56$
$\Rightarrow \frac{7 a}{4}=56$
$ \therefore a=32$
Then, numbers are
$ a =32$
$\Rightarrow a r=32 \times \frac{1}{2}=16$
$\Rightarrow a r^2=32 \times \frac{1}{4}=8$
$\Rightarrow 32,16,8 $
Hence, required numbers are $8,16,32$ or $32,16,8$.