Subjective Type

Between $$1$$ and $$31$$, $$m$$ numbers have been inserted in such a way that the resulting sequence is an $$A.P$$. and the ratio of $$\displaystyle { 7 }^{ th }$$ and $$(n-1)^{th}$$ numbers is $$5 : 9$$ . Find the value of $$n$$.

Solution

Let $$A_{ 1 },{ A }_{ 2 }...{ A }_{ n }$$ be n arithmetic means between $$1$$ and $$31$$

Then $$1,{ A }_{ 2, }{ A }_{ 3, }...{ A }_{ n, }{ 31 }$$ is an A.P with common difference $$\displaystyle d=\frac { 31-1 }{ n+1 } =\frac { 30 }{ n+1 } $$

Now $$\displaystyle { A }_{ 7 }=1+7d=1+\frac { 7\times 30 }{ n+1 } =\frac { n+211 }{ n+1 } $$

And $$\displaystyle { A }_{ n-1 }=1+\left( n-1 \right) d=1+\frac { 30 }{ n+1 } \left( n-1 \right) =\frac { 31n-29 }{ n+1 } $$

It is given that

$$\displaystyle \dfrac { { A }_{ 7 } }{ { A }_{ n-1 } } =\dfrac { 5 }{ 9 } $$

$$\Rightarrow \dfrac { n+211 }{ 31n-29 } =\dfrac { 5 }{ 9 } $$

$$\Rightarrow 9n+1899=155n-145$$

$$ \Rightarrow n=14$$

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