The \[{{24}^{th}}\] term of an A.P. is twice its \[{{10}^{th}}\] term. Show that its \[{{72}^{nd}}\] term is four times its \[{{15}^{th}}\] term.

The

${{24}^{th}}$

term of an A.P. is twice its

${{10}^{th}}$

term. Show that its

${{72}^{nd}}$

term is four times its

${{15}^{th}}$

term.

The

${{24}^{th}}$

term of an A.P. is twice its

${{10}^{th}}$

term. Show that its

${{72}^{nd}}$

term is four times its

${{15}^{th}}$

term.

From the question it is given that,

The

${{24}^{th}}$

term of an A.P. is twice its

${{10}^{th}}$

term =

${{a}_{24}}~=\text{ }2{{a}_{10}}$

We have to show that,

${{72}^{nd}}$

term is four times its

${{15}^{th}}$

term =

${{a}_{72}}~=\text{ }4{{a}_{15}}$

We know that,

$\begin{array}{*{35}{l}} {{a}_{24}}~=\text{ }a\text{ }+\text{ }23d\text{ }=\text{ }2{{a}_{10}} \\ a\text{ }+\text{ }23d\text{ }=\text{ }2\left( a\text{ }+\text{ }9d \right) \\ a\text{ }+\text{ }23d\text{ }=\text{ }2a\text{ }+\text{ }18d \\ 2a\text{ }\text{ }a\text{ }=\text{ }23d\text{ }\text{ }18d \\ \end{array}$

a = 5d … [equation (i)]

$\begin{array}{*{35}{l}} {{a}_{72}}~=\text{ }4{{a}_{15}} \\ a\text{ }+\text{ }71d\text{ }=\text{ }4\left( a\text{ }+\text{ }14d \right) \\ \end{array}$

Substitute the value of a we get,

$\begin{array}{*{35}{l}} 5d\text{ }+\text{ }71d\text{ }=\text{ }4\left( 5d\text{ }+\text{ }14d \right) \\ 76d\text{ }=\text{ }4\left( 19d \right) \\ \end{array}$

Therefore, it is proved that

${{72}^{nd}}$

term is four times its

${{15}^{th}}$

term

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