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If the \mathrm{p}^{\text {th }} and \mathrm{q}^{\text {th }} terms of a G.P. are q and p respectively, show that its (p+q)^{\text {th }} term is \left(\frac{\mathrm{q}^{\mathrm{p}}}{\mathrm{p}^{q}}\right)^{\frac{1}{\mathrm{p}-\mathrm{q}}}
CBSE | Class 11 | Maths | NCERT Exemplar | Sequences and Series
If the \mathrm{p}^{\text {th }} and \mathrm{q}^{\text {th }} terms of a G.P. are q and p respectively, show that its (p+q)^{\text {th }} term is \left(\frac{\mathrm{q}^{\mathrm{p}}}{\mathrm{p}^{q}}\right)^{\frac{1}{\mathrm{p}-\mathrm{q}}}

Solution:

The n^{\text {th }} term of Geometric Progression is given by t_{n}=a r^{n-1} in which the first term is a and the common difference is r

\mathrm{p}^{\text {th }} term is given as \mathrm{q}

\Rightarrow \mathrm{t}_{\mathrm{p}}=\mathrm{ar}^{\mathrm{p}-1}

Above equation can also be re-written as

\begin{array}{l} \Rightarrow \mathrm{q}=\mathrm{ar}^{\mathrm{p}-1} \\ \Rightarrow \mathrm{q}=\frac{\mathrm{ar}^{\mathrm{p}}}{\mathrm{r}} \end{array}

When we rearrange the above equation we obtain

\begin{array}{l} \Rightarrow \frac{\mathrm{a}}{\mathrm{r}}=\frac{\mathrm{q}}{\mathrm{r} p} \ldots(\mathrm{a}) \\ \mathrm{q}^{\text {th }} \text { term is given as } \mathrm{p} \\ \Rightarrow \mathrm{t}_{\mathrm{q}}=\mathrm{ar}^{\mathrm{r}-1} \\ \Rightarrow \mathrm{p}=\mathrm{ar}^{\mathrm{q}-1} \end{array}

Above equation can also be re-written as

\Rightarrow \mathrm{p}=\frac{\mathrm{ar}^{\mathrm{q}}}{\mathrm{r}}

When we rearrange the above equation we obtain

\Rightarrow \frac{\mathrm{a}}{\mathrm{r}}=\frac{\mathrm{p}}{\mathrm{r} \mathrm{q}} \ldots \text { (b) }

From eq.(a) and eq.(b) we have

\Rightarrow \frac{\mathrm{q}}{\mathrm{r}^{\mathrm{p}}}=\frac{\mathrm{p}}{\mathrm{rq}}

When we rearrange we obtain

\begin{array}{l} \Rightarrow \frac{\mathrm{q}}{\mathrm{p}}=\frac{\mathrm{r}^{\mathrm{p}}}{\mathrm{r}^{\mathrm{q}}} \\ \Rightarrow \mathrm{r}^{\mathrm{p}-\mathrm{q}}=\frac{\mathrm{q}}{\mathrm{p}} \end{array}

\begin{array}{l} \Rightarrow r^{p-q}=\frac{q}{p} \\ \Rightarrow r=\left(\frac{q}{p}\right)^{\frac{1}{p-q}} \end{array}

(p+q)^{\text {th }} term is given by

\begin{array}{l} \Rightarrow t_{p+q}=a r^{p+q-1} \\ \Rightarrow t_{p+q}=\left(a r^{p-1}\right) r^{q} \end{array}

But t_{p}=a r^{p-1} and the p^{\text {th }} term is q

\Rightarrow t_{p+q}=q r^{q}

But

\begin{array}{l} r=\left(\frac{q}{p}\right)^{\frac{1}{p-q}} \\ \Rightarrow t_{p+q}=q\left(\left(\frac{q}{p}\right)^{\frac{1}{p-q}}\right)^{q} \end{array}

By using the laws of exponents we obtain

\begin{array}{l} =q\left(\frac{q^{\frac{1}{p-q}}}{p^{\frac{1}{p-q}}}\right)^{q} \\ =q\left(\frac{q^{\frac{q}{p-q}}}{p^{\frac{q}{p-q}}}\right) \end{array}

On rearranging

=\frac{q_{1}{p^{-q}}+1}{p^{q\left(\frac{1}{p-q}\right)}}

Now take LCM and simplify it

\begin{array}{l} =\frac{q^{\frac{q+p-q}{p-q}}}{p^{q\left(\frac{1}{p-q}\right)}} \\ =\frac{q^{p\left(\frac{1}{p-q}\right)}}{p^{q\left(\frac{1}{p-q}\right)}} \\ \Rightarrow t_{p+q}=\left(\frac{q^{p}}{p^{q}}\right)^{\left(\frac{1}{p-q}\right)} \end{array}

As a result hence proved.

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