Question

Sequence and series question given in image solve please both question.
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Answer 1

$\large\underline{\sf{Solution-114}}$

We know, the general HP series is given by

$\rm :\longmapsto\:\boxed{ \tt{ \: \frac{1}{a}, \: \frac{1}{a + d}, \: \frac{1}{a + 2d}, - - - - \: }}$

and

↝ nᵗʰ term of an harmonic progression is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:= \frac{1}{\:a\:+\:(n\:-\:1)\:d} }}}}}} \\ \end{gathered}$

So, According to statement, the given HP series is

$\rm :\longmapsto\:2,\: 2\dfrac{1}{2},3 \dfrac{1}{3}, - - - -$

So, here,

$\rm :\longmapsto\:\dfrac{1}{a} = 2 \: \: \rm \implies\:a = \dfrac{1}{2}$

and

$\rm :\longmapsto\:\dfrac{1}{a + d} =2 \dfrac{1}{2}$

$\rm :\longmapsto\:\dfrac{1}{a + d} = \dfrac{5}{2}$

$\rm \implies\:a + d = \dfrac{2}{5}$

$\rm \implies\: \dfrac{1}{2} + d = \dfrac{2}{5}$

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

$\rm :\longmapsto\:d = \dfrac{4 - 5}{10}$

$\rm :\longmapsto\:d = \dfrac{ - 1}{10}$

$\bf\implies \:d = - \: \dfrac{1}{10}$

Hence,

$\rm :\longmapsto\:a_5 \: of \: HP \: is$

$\rm \: = \:\dfrac{1}{a + 4d}$

$\rm \: = \:\dfrac{1}{\dfrac{1}{2} - 4 \times \dfrac{1}{10} }$

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

$\rm \: = \:\dfrac{1}{\dfrac{5 - 4}{10} }$

$\rm \: = \:\dfrac{1}{\dfrac{1}{10} }$

$\rm \: = \:10$

$\rm \implies\:\boxed{ \tt{ \: \:a_5 \: of \: HP \: = \: 10 \: }}$

• Hence, Option (d) is correct.

$\red{\large\underline{\sf{Solution-115}}}$

According to statement,

$\rm :\longmapsto\:a_7 = \dfrac{1}{10}$

$\rm :\longmapsto\:\dfrac{1}{a + 6d} = \dfrac{1}{10}$

$\rm \implies\:a + 6d = 10 - - - - (1)$

Also, given that

$\rm :\longmapsto\:a_{12} = \dfrac{1}{25}$

$\rm :\longmapsto\:\dfrac{1}{a + 11d} = \dfrac{1}{25}$

$\rm \implies\:a + 11d = 25 - - - - (2)$

On Subtracting equation (1) from equation (2), we get

$\rm :\longmapsto\:5d = 15$

$\bf\implies \:d \: = \: 3$

On substituting d = 3, in equation (1), we get

$\rm :\longmapsto\:a + 6d = 10$

$\rm :\longmapsto\:a + 6(3) = 10$

$\rm :\longmapsto\:a + 18 = 10$

$\bf\implies \:a \: = \: - \: 8$

Hence,

$\rm :\longmapsto\:a_{20} \: of \: HP \: is$

$\rm \: = \:\dfrac{1}{a + 19d}$

$\rm \: = \:\dfrac{1}{ - 8 + 19(3)}$

$\rm \: = \:\dfrac{1}{ - 8 + 57}$

$\rm \: = \:\dfrac{1}{49}$

Thus,

$\rm \implies\:\boxed{ \tt{ \: \:a_{20} \: of \: HP \: = \: \frac{1}{49} \: }}$

• So, option (d) is correct

Additional Information :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

↝ nᵗʰ term of an geometric sequence is,

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

↝ nᵗʰ term of an harmonic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:= \frac{1}{\:a\:+\:(n\:-\:1)\:d} }}}}}} \\ \end{gathered}$