From the question it is given that,
a4 = x
a7 = y
a10 = z
Now we have to prove that, x, y, z are in G.P.
Then, by the formula
![\[{{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5c4a26b205addff26274971977b92900_l3.png "Rendered by QuickLaTeX.com")
![\[\begin{array}{*{35}{l}} {{a}_{4}}~=\text{ }a{{r}^{4\text{ }\text{ }1}} \\ {{a}_{4}}~=\text{ }{{a}^{3}} \\ {{a}_{4}}~=\text{ }x \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-463aea1962435f49bb63772f10898d44_l3.png "Rendered by QuickLaTeX.com")
So,
![\[{{a}_{7}}~=\text{ }{{a}^{7\text{ }\text{ }1}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-30add35098340bada99a135408486c11_l3.png "Rendered by QuickLaTeX.com")
![\[\begin{array}{*{35}{l}} {{a}_{7}}~=\text{ }{{a}^{6}} \\ {{a}_{7}}~=\text{ }y \\ {{a}_{10}}~=\text{ }{{a}^{10\text{ }-1}} \\ {{a}_{10}}~=\text{ }{{a}^{9}} \\ {{a}_{10}}~=\text{ }z \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7d421c1b9e4764dbbf08f97b84d06680_l3.png "Rendered by QuickLaTeX.com")
x, y, z are in G.P. then,
y2 = xz
Then, xz =
![\[a{{r}^{3}}~\times \text{ }a{{r}^{9}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e18a0c406781d681871a6591c3755245_l3.png "Rendered by QuickLaTeX.com")
![\[\begin{array}{*{35}{l}} =\text{ }{{a}^{2}}{{r}^{3}}{{~}^{+\text{ }9}} \\ =\text{ }{{a}^{2}}{{r}^{12}} \\ {{y}^{2}}~=\text{ }{{(a{{r}^{6}})}^{2}} \\ {{y}^{2}}~=\text{ }{{a}^{2}}{{r}^{12}} \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-65c0276f86ec97ba815b1c73dd276092_l3.png "Rendered by QuickLaTeX.com")
By comparing left hand side and right hand side
LHS = RHS
Therefore, x, y and z are in G.P.