Question

If 1, x, y, z, 1296/2401 are in G.P. then find the value of x+y+z and the common ratio of the G.P.

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Answer 1

Given : 1, x, y, z, 1296/2401 are in G.P

To find : The value of x+y+z and the common ratio of the G.P

Solution :

Let's assume that a be the first term and r be the common ratio of the given GP.

We have,

• The first term of the GP = a = 1
• The 5th term of the GP = a₅ = 1296/2401

nth term of any GP is given by,

• an = ar⁽ⁿ⁻¹⁾

So by using this general form, 5th term of the GP is given by,

• a₅ = ar⁴

On substituting the given values,

⇒ 1296/2401 = r⁴

⇒ 6⁴/7⁴ = r⁴

⇒ (6/7)⁴ = r⁴

⇒ ± 6/7 = r

Therefore two cases arisen :-

Case I - When r = 6/7

• x = ar = 6/7
• y = ar² = 36/49
• z = ar³ = 216/343

Case II - When r = -6/7

• x = ar = -6/7
• y = ar² = 36/49
• z = ar³ = -216/343

Required result for case I :-

⇒ x + y + z

⇒ 6/7 + 36/49 + 216/343

⇒ (294+252+216)/343

⇒ 762/343

Required result for case II :-

⇒ x + y + z

⇒ - 6/7 + 36/49 - 216/343

⇒ (- 294+252 - 216)/343

⇒ -258/343

Answer :-

• Common ratio = ± 6/7
• x + y + z = 762/343 or -258/343

Answer 2

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

Given that,

$\red{\rm :\longmapsto\:1,x,y,z,\dfrac{1296}{2401} \: are \: in \: GP}$

Let assume that the common ratio of GP series is r.

So, we have

$\rm :\longmapsto\:a_1 = 1$

$\rm :\longmapsto\:a_5 = \dfrac{1296}{2401}$

We know that,

↝ nᵗʰ term of an geometric sequence is,

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

where,

a is the first term of GP series

r is the common ratio

n is number of terms

Thus,

$\rm :\longmapsto\:a_5 = \dfrac{1296}{2401}$

$\rm :\longmapsto\:a {r}^{5 - 1} = \dfrac{1296}{2401}$

$\rm :\longmapsto\:1 \times {r}^{4} = \dfrac{1296}{2401}$

$\rm :\longmapsto\: {r}^{4} = \dfrac{6 \times 6 \times 6 \times 6}{7 \times 7 \times 7 \times 7}$

$\rm :\longmapsto\: {r}^{4} = {\bigg[\dfrac{6}{7} \bigg]}^{4}$

$\bf\implies \:r \: = \: \pm \: \dfrac{6}{7}$

So, two cases arises

Case - 1

$\rm :\longmapsto\:When \: r = \dfrac{6}{7}$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = ar = \dfrac{6}{7} } \\ \\ &\sf{y = {ar}^{2} = \dfrac{36}{49} }\\ \\ &\sf{z = {ar}^{3} = \dfrac{216}{343} } \end{cases}\end{gathered}\end{gathered}$

Thus,

$\rm :\longmapsto\:x + y + z$

$\rm \:  =  \:\dfrac{6}{7} + \dfrac{36}{49} + \dfrac{216}{343}$

$\rm \:  =  \:\dfrac{294 + 252 + 216}{343}$

$\rm \:  =  \:\dfrac{762}{343}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \: x + y + z  =  \:\dfrac{762}{343} \: \: }}$

Case :- 2

$\rm :\longmapsto\:When \: r \: = - \: \dfrac{6}{7}$

So,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{x = ar \: = - \: \dfrac{6}{7} } \\ \\ &\sf{y = {ar}^{2} = \dfrac{36}{49} }\\ \\ &\sf{z = {ar}^{3} \: = - \: \dfrac{216}{343} } \end{cases}\end{gathered}\end{gathered}$

Thus,

$\rm :\longmapsto\:x + y + z$

$\rm \:  =  \: - \: \dfrac{6}{7} + \dfrac{36}{49} - \dfrac{216}{343}$

$\rm \:  =  \:\dfrac{ - 294 + 252 - 216}{343}$

$\rm \:  =  \:\dfrac{ - 258}{343}$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \: x + y + z  =  \: - \: \dfrac{258}{343} \: \: }}$