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Home » The     terms of a G.P. are p, q and s, respectively. Show that     .
The

    \[{{\mathbf{5}}^{\mathbf{th}}},\text{ }{{\mathbf{8}}^{\mathbf{th}}}~\mathbf{and}\text{ }\mathbf{1}{{\mathbf{1}}^{\mathbf{th}}}~\]

terms of a G.P. are p, q and s, respectively. Show that

    \[\mathbf{q}{}^\text{2}\text{ }=\text{ }\mathbf{ps}\]

.
Arithmetic and Geometric Progression | Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal
The

    \[{{\mathbf{5}}^{\mathbf{th}}},\text{ }{{\mathbf{8}}^{\mathbf{th}}}~\mathbf{and}\text{ }\mathbf{1}{{\mathbf{1}}^{\mathbf{th}}}~\]

terms of a G.P. are p, q and s, respectively. Show that

    \[\mathbf{q}{}^\text{2}\text{ }=\text{ }\mathbf{ps}\]

.

From the question it is given that,

    \[\begin{array}{*{35}{l}} {{a}_{5}}~=\text{ }p \\ {{a}_{8}}~=\text{ }q \\ {{a}_{11}}~=\text{ }s \\ \end{array}\]

Now we have to prove that,

    \[\mathbf{q}{}^\text{2}\text{ }=\text{ }\mathbf{ps}\]

Then, by the formula

    \[{{a}_{n}}~=\text{ }a{{r}^{n\text{ }\text{ }1}}\]

    \[\begin{array}{*{35}{l}} {{a}_{5}}~=\text{ }a{{r}^{5\text{ }\text{ }1}} \\ {{a}_{5}}~=\text{ }{{a}^{4}} \\ {{a}_{5}}~=\text{ }p \\ \end{array}\]

So,

    \[~{{a}_{8}}~=\text{ }{{a}^{8\text{ }\text{ }1}}\]

    \[\begin{array}{*{35}{l}} {{a}_{8}}~=\text{ }{{a}^{7}} \\ {{a}_{8}}~=\text{ }q \\ {{a}_{11}}~=\text{ }{{a}^{11\text{ }-1}} \\ {{a}_{11}}~=\text{ }{{a}^{10}} \\ {{a}_{11}}~=\text{ }s \\ \end{array}\]

p, q, s are in G.P. then,

    \[\begin{array}{*{35}{l}} {{q}^{2}}~=\text{ }{{(a{{r}^{7}})}^{2}} \\ =\text{ }a{{r}^{14}} \\ \end{array}\]

Then,

    \[px\text{ }=\text{ }a{{r}^{4}}~\times \text{ }a{{r}^{10}}\]

    \[\begin{array}{*{35}{l}} =\text{ }{{a}^{2}}{{r}^{4}}{{~}^{+\text{ }10}} \\ =\text{ }{{a}^{2}}{{r}^{14}} \\ \end{array}\]

Therefore,

    \[{{q}^{2}}~=\text{ }ps\]

i

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