Let {A_n} = left( {frac{3}{4}} right) - {left | vedclass

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Question

${A_n} = \left( {\frac{3}{4}} \right) - {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} - ...... + {\left( { - 1} \right)^{n - 1}}

Vedclass · Accepted Answer

$7$

Vedclass · Answer

${A_n} = \left( {\frac{3}{4}} ight) - {\left( {\frac{3}{4}} ight)^2} + {\left( {\frac{3}{4}} ight)^3} - ...... + {\left( { - 1} ight)^{n - 1}}{\left( {\frac{3}{4}} ight)^n}$ Which is a $G.P.$ with $a = \frac{3}{4}'r = \frac{{ - 3}}{4}$ and number of terms $=n$ ${A_n} = \frac{{\frac{3}{4} imes \left( {1 - {{\left( {\frac{{ - 3}}{4}} ight)}^n}} ight)}}{{1 - \left( {\frac{{ - 3}}{4}} ight)}} - \frac{{\frac{3}{4} imes \left( {1 - {{\left( {\frac{{ - 3}}{4}} ight)}^n}} ight)}}{{\frac{7}{4}}}$ $ \Rightarrow {A_n} = \frac{3}{7}\left[ {1 - {{\left( {\frac{{ - 3}}{4}} ight)}^n}} ight]\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 ight)$ As, ${B_n} = 1 - {A_n}$ For least odd natural number $p$, such that ${B_n} > {A_n}$ $ \Rightarrow 1 - {A_n} > {A_n}\,\,\,\,\, \Rightarrow 1 > 2 > imes {A_n}\,\,\,\,\, \Rightarrow {A_n} < \frac{1}{2}$ From eqn. $(1)$, we get $\frac{3}{7} imes \left[ {1 - {{\left( {\frac{{ - 3}}{4}} ight)}^n}} ight] < \frac{1}{2}\,\, \Rightarrow 1 - {\left( {\frac{{ - 3}}{4}} ight)^n} < \frac{7}{6}$ $ \Rightarrow 1 - \frac{7}{6} < {\left( {\frac{{ - 3}}{4}} ight)^n} \Rightarrow \frac{{ - 1}}{6} < {\left( {\frac{{ - 3}}{4}} ight)^n}$ As $n$ is odd, then ${\left( {\frac{{ - 3}}{4}} ight)^n} = - \frac{{{3^n}}}{4}$ So $\frac{{ - 1}}{6} < - {\left( {\frac{3}{4}} ight)^n}\,\,\, \Rightarrow \frac{1}{6} > {\left( {\frac{3}{4}} ight)^n}$ $\log \left( {\frac{1}{6}} ight) = n\,\log \left( {\frac{3}{4}} ight) \Rightarrow 6.228 < n$ Hence, $n$ should be $7$.

Let ${A_n} = \left( {\frac{3}{4}} \right) - {\left( {\frac{3}{4}} \right)^2} + {\left( {\frac{3}{4}} \right)^3} - ..... + {\left( { - 1} \right)^{n - 1}}{\left( {\frac{3}{4}} \right)^n}$ and $B_n \,= 1 - A_n$ . Then, the least odd natural number $p$ , so that ${B_n} > {A_n}$, for all $n \geq p$ is

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