Question

If a, b, c, d are in GP, prove that

$(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd{)}^2$

Answer 1

Let r be the common ratio of the GP a, b, c, d.

Then, b = ar, $c=a{r}^2$ and $d=a{r}^3$

$\therefore$ $LHS=(a^2+b^2+c^2)(b^2+c^2+d^2)$

$=(a^2+a^2{r}^2+a^2{r}^4)(a^2{r}^2+a^2{r}^4+a^2{r}^6)$

$=a^4{r}^2(1+r^2+r^4{)}^2$

And, $RHS=(ab+bc+cd{)}^2=(a^{2}r+a^2{r}^3+a^2{r}^5{)}^2$

$a^4{r}^2(1+r^2+r^4{)}^2$

Hence, $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd{)}^2$