Question

Solve previous year Question of iit jee

Chapter :- sequence and series​

Answer 1

Answer:

let the three positive numbers forming GP be a, ar , ar².

Now,

Middle term of GP is doubled,,

a, 2ar , ar² are in AP.

$→2ar = \frac{a + {ar}^{2} }{2}$

$→4ar = a(1 + {r}^{2} )$

$→4r = 1 + {r}^{2}$

$→ {r}^{2} - 4r + 1 = 0$

By the quadratic formula :-

$r \: = \frac{ - ( - 4)± \sqrt{ {( - 4)}^{2} - 4 \times 1 \times 1} }{2 \times 1}$

$→r = \frac{ 4 \:± \sqrt{16 - 4} }{2}$

$→r = \frac{4 \:± \sqrt{12} }{2}$

$→r = \frac{4±2 \sqrt{3} }{2}$

$→r = \frac{2(2± \sqrt{3} )}{2}$

So,,

$r = 2± \sqrt{3}$

Acc to ques, The GP is increasing so we will take 'r' as positive value,

hence, the common ratio of GP, r = 2+√3

Answer 2

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

Let suppose that 3 numbers in GP series as

$\red{\rm :\longmapsto\:a,\: ar, \: {ar}^{2} \: such \: that \: r > 1}$

Now, it is further given that, if middle term is doubled the numbers are in AP.

$\red{\rm :\longmapsto\:a,\: 2ar, \: {ar}^{2} \: are \: in \: AP}$

We know, in AP series, common difference us same

$\red{\rm :\longmapsto\:\: 2ar - a = \: {ar}^{2} - 2ar}$

$\rm :\longmapsto\: {ar}^{2} + a = 4ar$

On cancel out a, we get

$\rm :\longmapsto\: {r}^{2} + 1 = 4r$

$\rm :\longmapsto\: {r}^{2} - 4r + 1 = 0$

Its a quadratic in r, so to find out the value of r, we use Quadratic Formula,

$\rm :\longmapsto\:r = \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

Here,

$\rm :\longmapsto\:a = 1$

$\rm :\longmapsto\:b = - 4$

$\rm :\longmapsto\:c = 1$

Thus,

$\rm :\longmapsto\:r = \dfrac{ - ( - 4) \: \pm \: \sqrt{ {( - 4)}^{2} - 4(1)(1)} }{2(1)}$

$\rm :\longmapsto\:r = \dfrac{4 \: \pm \: \sqrt{ 16 - 4} }{2}$

$\rm :\longmapsto\:r = \dfrac{4 \: \pm \: \sqrt{ 12} }{2}$

$\rm :\longmapsto\:r = \dfrac{4 \: \pm \: \sqrt{ 2 \times 2 \times 3} }{2}$

$\rm :\longmapsto\:r = \dfrac{4 \: \pm \: 2\sqrt{3} }{2}$

$\rm :\longmapsto\:r = \: 2 \pm \: \sqrt{3}$

$\bf :\implies\:r = 2 + \sqrt{3} \: \: as \: r > 1$

Hence,

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

Additional Information :-

For Geometric Progression,

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

$\red{\rm :\longmapsto\:S_n = \dfrac{a( {r}^{n} - 1)}{r - 1} \: \: provided \: that \: r \ne \: 1}$