Question

There are n arithmetic means between 9 and 27. If the ratio of the last mean to the first mean is 2:1, Find the value of n​

Answer 1

$\huge{\tt{\underline{ GIVEN }}} \::$

• $\Large\textsf{There are n arithmetic means between 9 and 27.}$
• $\Large\textsf{The ratio of the last mean to the first mean is 2:1}$

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$\huge{\tt{\underline{TO \ FIND}}} \::$

• $\large\textsf{The value of n}$

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$\huge{\tt{\underline{SOLUTION}}} \::$

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$\large\Rightarrow\:\:\textsf{a + (n + 2 - 1) d = 27}$

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$\large\Rightarrow\:\:\textsf{9 + (n + 1) d = 27}$

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$\large\Rightarrow\:\:\textsf{(n + 1) d = 27 - 9}$

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$\large\Rightarrow\:\:\textsf{(n + 1) d = 18}$

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$\large\Rightarrow\:\:{\boxed{\sf{d = \dfrac{18}{n + 1}}}}$

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$\large\star\:\:\sf{\dfrac{a_{n}}{a_{1}} = \dfrac{T_{n+1}}{T_{2}} = \dfrac{2}{1}}$

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$\large\longrightarrow\:\:\sf{\dfrac{a + (n + \cancel{1 - 1}) d}{a + (2 - 1) d} = \dfrac{2}{1}}$

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$\large\longrightarrow\:\:\sf{\dfrac{a + (n) d}{a + d} = \dfrac{2}{1}}$

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$\large\longrightarrow\:\:\sf{\dfrac{9 + (n) \dfrac{18}{(n+1)}}{9 + \dfrac{18}{(n+1)}} = \dfrac{2}{1}}$

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$\large\longrightarrow\:\:\sf{\dfrac{\dfrac{9n + 9 + 18n}{(n + 1)}}{\dfrac{9n + 9 + 18}{(n + 1)}} = \dfrac{2}{1}}$

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$\large\longrightarrow\:\:\sf{\dfrac{27n + 9}{9n + 27} = \dfrac{2}{1}}$

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$\large\longrightarrow\:\:\sf{ 27n + 9 = 18n + 54}$

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$\large\longrightarrow\:\:\sf{9n = 45}$

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$\large\longrightarrow\:\:\sf{n = \dfrac{\cancel{45}{\Large{^{\:\:5}}}}{\cancel{9}{\Large{_{\:\:1}}} } }$

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$\large\longrightarrow\:\:\sf{n = 5}$