Question

The numbers $a,b,c$ and $d$ form a geometric, progression. Find $(a-c{)}^2+(b-c{)}^2+(b-d{)}^2-(a-d{)}^2$

Answer 1

${(a-c)}^2+{(b-c)}^2+{(b-d)}^2-{(a-d)}^2$

As,$a,b,c\&d$ form geometric progression, Let$a=a_1,b=a_{1}r,c=a_1{r}^2$ and $d=a_1{r}^3$

$\therefore {(a-c)}^2+{(b-c)}^2+{(b-d)}^2-{(a-d)}^2=a^2-2ac+c^2+b^2-2bc+c^2+b^2-2bd+d^2-a^2+2ad-d^2=2b^2+2c^2-2ac-2bc-2bd+2ad=2{(b-c)}^2-2(ac+bd-ad)$

Replacing the values of $a,b,c\&d$

$=2{[a_{1}r-a_1{r}^2]}^2-2[a_1^2{r}^2+a_1^2{r}^4-a_1^2{r}^3]=2a_1^2{[r-r^2]}^2-2a_1^2[r^2+r^4-r^3]=2a_1^2[{(r-r^2)}^2-r^2-r^4+r^3]=2a_1^2[r^2-2r^3+r^4-r^2-r^4+r^3]=2a_1^2\times (-r^3)=-2a_1^2{r}^3$