Question

plz ans this 11th ncert sequence and series question​

Answer 1

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

Given that,

$\rm :\longmapsto\:a_1,a_2,a_3, - - - ,a_{n - 1},a_n \: are \: in \: AP$

$\bf\implies \:a_1 + a_n = a_2 + a_{n - 1} = a_3 + a_{n - 2} = - - -$

Further given that,

$\rm :\longmapsto\:\dfrac{1}{a_1a_n} + \dfrac{1}{a_2a_{n - 1}} + \dfrac{1}{a_3a_{n - 2}} + - - + \dfrac{1}{a_na_1} \\ \rm \: = \lambda \: \bigg[\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3} + - - - +\dfrac{1}{a_{n - 1}} + \dfrac{1}{a_n} \bigg]$

Now, Consider,

$\rm :\longmapsto\:\dfrac{1}{a_1a_n} + \dfrac{1}{a_2a_{n - 1}} + \dfrac{1}{a_3a_{n - 2}} + - - + \dfrac{1}{a_na_1}$

Can be rewritten as

$\rm = \:\dfrac{1}{a_1 + a_n} \bigg[\dfrac{a_1 + a_n}{a_1a_n} + \dfrac{a_1 + a_n}{a_2a_{n - 1}} + \dfrac{a_1 + a_n}{a_3a_{n - 2}} + - - + \dfrac{a_1 + a_n}{a_na_1}\bigg]$

$\rm = \:\dfrac{1}{a_1 + a_n} \bigg[\dfrac{a_1 + a_n}{a_1a_n} + \dfrac{a_2 + a_{n - 1}}{a_2a_{n - 1}} + \dfrac{a_3 + a_{n - 2}}{a_3a_{n - 2}} + - - + \dfrac{a_n + a_1}{a_na_1}\bigg]$

can be further rewritten as

$\rm=\dfrac{1}{a_1 + a_n}\bigg[\dfrac{1}{a_1} + \dfrac{1}{a_n} + \dfrac{1}{a_2} + \dfrac{1}{a_{n - 1}} + - - - + \dfrac{1}{a_1} + \dfrac{1}{a_n} \bigg]$

$\rm=\dfrac{2}{a_1 + a_n}\bigg[\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3} + - - - + \dfrac{1}{a_{n - 1}} + \dfrac{1}{a_n} \bigg]$

Now, as it is given that,

$\red{\rm :\longmapsto\:\dfrac{1}{a_1a_n} + \dfrac{1}{a_2a_{n - 1}} + \dfrac{1}{a_3a_{n - 2}} + - - + \dfrac{1}{a_na_1}} \\ \rm \: = \red{ \lambda \: \bigg[\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3} + - - - +\dfrac{1}{a_{n - 1}} + \dfrac{1}{a_n} \bigg]}$

So, we concluded that,

$\red{\bf\implies \:\boxed{ \rm{ \: \lambda \: = \: \dfrac{2}{a_1 + a_n} \: \: }}}$

Hence, Option (4) is correct.