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