Question

If the $4^{th},{10}^{th}and{16}^{th}$ terms of a G.P, are x, y and z respectively. Prove that x, y, z are in G.P

Answer 1

Let 'a' be the first term and 'r' be the common ratio of the given G.P.

Here, $a_4=x\Rightarrow a{r}^3=x......\left(1\right)$

$a_{10}=y\Rightarrow a{r}^9=y......\left(2\right)$

$a_{16}=z\Rightarrow a{r}^{15}=z......\left(3\right)$

From (2), we have :

$[a{r}^9{]}^2=y^2\Rightarrow y^2=a^2{r}^{18}$

$\Rightarrow y^2=\left(a{r}^3\right)\left(a{r}^{15}\right)$

$\Rightarrow y^2=xz[\therefore n=a{r}^{3}andz=a{r}^{15}]$

Which shows that x, y, z are in G.P.