Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

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

Given that,

m is Arithmetic mean between l and n,

It means l, m, n are in AP

So,

$\rm :\longmapsto\:m = \dfrac{l + n}{2} - - - (1)$

Also, given that,

$\rm :\longmapsto\:G_1,G_2,G_3 \: are \: geometric \: mean \: between \: l \: and \: n$

$\rm :\longmapsto\:l,G_1,G_2,G_3,n \: are \: in \: GP$

We know,

nth term of GP series is given by

$\rm :\longmapsto\:a_n = {ar}^{n - 1}$

where

• a is the first term

• r is common ratio

• n is number of terms

So,

According to given,

• first term = l

• last term = n

• number of terms, = 5

On substituting the values, we have

$\rm :\longmapsto\:n = l {r}^{5 - 1}$

$\rm :\longmapsto\:n = l {r}^{4}$

$\bf\implies \: {r}^{4} = \dfrac{n}{l} - - - (2)$

Now,

$\rm :\longmapsto\:G_1 = lr$

$\rm :\longmapsto\:G_2 = l {r}^{2}$

$\rm :\longmapsto\:G_3 = l {r}^{3}$

Now, Consider

$\rm :\longmapsto\: {G_1}^{4} + 2( {G_2}^{4}) + {G_3}^{4}$

On substituting the values we get

$\rm \: = \: {(lr)}^{4} + 2 {( {lr}^{2}) }^{4} + {( {lr}^{3} )}^{4}$

$\rm \: = \: \: {l}^{4} {r}^{4}(1 + 2{r}^{4} + {r}^{8})$

$\rm \: = \: \: {l}^{4} {r}^{4} {(1 + {r}^{4} )}^{2}$

$\rm \: = \: \: {l}^{4} \times \dfrac{n}{l} \bigg(1 + \dfrac{n}{l} \bigg)^{2}$

$\rm \: = \: \: {l}^{3} n \bigg( \dfrac{l + n}{l} \bigg)^{2}$

$\rm \: = \: \: {l}^{3} n \bigg( \dfrac{2m}{l} \bigg)^{2}$

$\rm \: = \: \: {l}^{3} n \bigg( \dfrac{4 {m}^{2} }{ {l}^{2} } \bigg)$

$\rm \: = \: \: {4m}^{2}ln$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf \: Option \: (b) \: is \: correct}}$