Question

i want it's detailed solutions​

Answer 1

Answer:

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Answer 2

$\large\underline{\sf{Solution-}}$

The given infinite series is

$\rm :\longmapsto\:\dfrac{2}{5} + \dfrac{3}{ {5}^{2} }+\dfrac{2}{ {5}^{3} } + \dfrac{3}{ {5}^{4} }+\dfrac{2}{ {5}^{5} } + \dfrac{3}{ {5}^{6}} + - - - - \infty$

The above can be rewritten as

$\rm \: = \:\bigg(\dfrac{2}{5} + \dfrac{2}{ {5}^{3} }+\dfrac{2}{ {5}^{5} } + - - \infty \bigg) +\bigg( \dfrac{3}{ {5}^{2} }+\dfrac{3}{ {5}^{4} } + \dfrac{3}{ {5}^{6}} + - - \infty\bigg)$

$\rm \:  =  \:  \:z_1 + z_2$

Where,

$\rm :\longmapsto\: \: z_1 = \:\bigg(\dfrac{2}{5} + \dfrac{2}{ {5}^{3} }+\dfrac{2}{ {5}^{5} } + - - \infty \bigg)$

and

$\rm :\longmapsto\:z_2 \: = \: \bigg( \dfrac{3}{ {5}^{2} }+\dfrac{3}{ {5}^{4} } + \dfrac{3}{ {5}^{6}} + - - \infty\bigg)$

Consider,

$\rm :\longmapsto\: \: z_1 = \dfrac{2}{5} + \dfrac{2}{ {5}^{3} }+\dfrac{2}{ {5}^{5} } + - - \infty$

Its an infinite GP series with

$\rm :\longmapsto\:a = \dfrac{2}{5}$

$\rm :\longmapsto\:r = \dfrac{a_2}{a_1} = \dfrac{1}{ {5}^{2} }$

We know,

Sum of infinite GP series is given by

$\boxed{ \sf{ \:S_ \infty = \dfrac{a}{1 - r} \: provided \: that \: - 1 < r < 1}}$

So, on substituting the values of a and r, we get

$\rm :\longmapsto\:z_1 = \dfrac{\dfrac{2}{5} }{1 - \dfrac{1}{ {5}^{2} } }$

$\rm :\longmapsto\:z_1 = \dfrac{\dfrac{2}{5} }{1 - \dfrac{1}{ {25}} }$

$\rm :\longmapsto\:z_1 = \dfrac{\dfrac{2}{5} }{\dfrac{25 - 1}{ {25}} }$

$\rm :\longmapsto\:z_1 = \dfrac{\dfrac{2}{5} }{\dfrac{24}{ {25}} }$

$\bf\implies \:z_1 = \dfrac{5}{12}$

Now,

Consider,

$\rm :\longmapsto\:z_2 \: = \: \dfrac{3}{ {5}^{2} }+\dfrac{3}{ {5}^{4} } + \dfrac{3}{ {5}^{6}} + - - \infty$

Its an Infinite GP series with

$\rm :\longmapsto\:a = \dfrac{3}{ {5}^{2} }$

$\rm :\longmapsto\:r = \dfrac{a_2}{a_1} = \dfrac{1}{ {5}^{2} }$

So, on applying the formula of sum of infinite GP, we get

$\rm :\longmapsto\:z_2 = \dfrac{\dfrac{3}{ {5}^{2} } }{1 - \dfrac{1}{ {5}^{2} } }$

$\rm :\longmapsto\:z_2 = \dfrac{\dfrac{3}{ {5}^{2} } }{1 - \dfrac{1}{ {25}} }$

$\rm :\longmapsto\:z_2 = \dfrac{\dfrac{3}{ {5}^{2} } }{\dfrac{25 - 1}{ {25}} }$

$\rm :\longmapsto\:z_2 = \dfrac{\dfrac{3}{ {5}^{2} } }{\dfrac{24}{ {25}} }$

$\bf\implies \:z_2 = \dfrac{1}{8}$

Hence,

$\rm :\longmapsto\:\dfrac{2}{5} + \dfrac{3}{ {5}^{2} }+\dfrac{2}{ {5}^{3} } + \dfrac{3}{ {5}^{4} }+\dfrac{2}{ {5}^{5} } + \dfrac{3}{ {5}^{6}} + - - - - \infty$

$\rm \: = \:\bigg(\dfrac{2}{5} + \dfrac{2}{ {5}^{3} }+\dfrac{2}{ {5}^{5} } + - - \infty \bigg) +\bigg( \dfrac{3}{ {5}^{2} }+\dfrac{3}{ {5}^{4} } + \dfrac{3}{ {5}^{6}} + - - \infty\bigg)$

$\rm \:  =  \:  \:z_1 + z_2$

$\rm \:  =  \:  \:\dfrac{5}{12} + \dfrac{1}{8}$

$\rm \:  =  \:  \:\dfrac{10 + 3}{24}$

$\rm \:  =  \:  \:\dfrac{13}{24}$

Hence,

• Option (c) is correct.

Additional Information :-

↝ nᵗʰ term of a geometric sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\: {r}^{ \: n \: - \: 1 \: } }}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.

↝ Sum of first n terms of geometric sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{a( {r}^{n} - 1)}{r - 1} \: provided \: that \: r \ne \: 1}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• Sₙ is the sum of n terms of GP.

• a is the first term of the sequence.

• n is the no. of terms.

• r is the common ratio.