Question
If $m$ arithmetic means $( A . Ms )$ and three geometric means $(G.Ms)$ are inserted between $3$ and $243$ such that $4^{\text {th }}$ $A.M.$ is equal to $2^{\text {nd }}$ $G.M.$, then $m$ is equal to
$d =\frac{243-3}{ m +1}=\frac{240}{ m +1}$
Now $3, G _{1}, G _{2}, G _{3}, 243$
$r=\left(\frac{243}{3}\right)^{\frac{1}{3+1}}=3$
$\therefore \quad A_{4}=G_{2}$
$\Rightarrow \quad a +4 d = ar ^{2}$
$3+4\left(\frac{240}{ m +1}\right)=3(3)^{2}$
$m=39$
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$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{ e } x d x=\frac{1}{\alpha}\left(\frac{ x }{ e }\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{ e }{ x }\right)^{\delta x }+ C ,$
Where $e =\sum \limits_{ n =0}^{\infty} \frac{1}{ n !}$ and $C$ is constant of integration, then $\alpha+2 \beta+3 \gamma-4 \delta$ is equal to: