Question

Three integers $a,b,c$ are in $G.P.$ If $a,b,c-64$ are in $A.P.$, and $a,b-8,c-64$ are in $G.P.$, then $(a+b+c)$ is equal to

• A. $124$
• B. $214$
• C. $412$
• D. $142$

Answer 1

The correct option is A $124$

$a,b,c$ are in $G.P$. with common ratio $r.$
$\Rightarrow b=ar,c=a{r}^2$

$a,b,c-64$ are in $A.P.$
$\Rightarrow a+c-64=2b$
$\Rightarrow a+a{r}^2-64=2ar$
$\Rightarrow a(r^2-2r+1)=64...\left(1\right)$

$a,b-8,c-64$ are in $G.P.$
$\Rightarrow (b-8{)}^2=a(c-64)$
$\Rightarrow (ar-8{)}^2=a(a{r}^2-64)$
$\Rightarrow a(16r-64)=64...\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$r^2-2r+1=16r-64$
$\Rightarrow r^2-18r+65=0\Rightarrow r=5\text{or}r=13$
$\Rightarrow a=4\text{or}a=\frac{4}{9}$ (not applicable as $a,b,c$ are integers)
$\therefore a=4,b=20,c=100\Rightarrow a+b+c=124$