Question

the product of 4 numbers in GP is 64 the ratio of the sum of the means to the sum of the extremes is 2/3 find the numbers​

Answer 1

Answer:

numbers are  1 , 2 4  8

or -1 , -2 , -4 , -8

Step-by-step explanation:

Let say 4 numbers are

a , ar  , ar² , ar³  

a = first term

r = common ratio

the product of 4 numbers in GP is 64

=> a * ar * ar² * ar³   = 64

=> a⁴ r⁶ = 64

(ar + ar²)/(a + ar³)  = 2/3

=> 3r + 3r² = 2 + 2r³

=> 2r³ - 3r² - 3r + 2 = 0

=> (r - 2)(2r² + r  - 1) = 0

=> (r - 2) (2r² + 2r - r - 1) = 0

=> (r - 2)( 2r(r + 1) - 1(r + 1)) = 0

=> (r - 2)(2r - 1)(r + 1) = 0

=> r = 2  , 1/2  , -1    but r = -1 make denominator 0 so not possible

putting r = 2

a⁴ r⁶ = 64   => a⁴2⁶ = 64 => a⁴ = 1  => a = ± 1

numbers are

1 , 2 , 4 , 8   or   -1 , -2 , -4 , - 8

putting r = 1/2

a⁴ r⁶ = 64   => a⁴(1/2)⁶ = 64 => a⁴ = 64 * 64  => a = ± 8

numbers are

8 , 4 , 2 , 1   or  -8 - 4  , -2 , -1

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

First term be t

Common ratio be c

Hence,

$\textbf{\underline{Numbers\;are :- }}$

t , tc , tc² , tc³

${\boxed{\sf\:{Product\;of\;four\;numbers\;in\;GP = 64 }}}$

⇒ t × tc × tc² × tc³ = 64

⇒ t⁴(c)^⁶ = 64

$\tt{\rightarrow\dfrac{(tc+tc^2)}{(t+tc^3)}=\dfrac{2}{3}}$

⇒ 3c + 3c² = 2 + 2c³

⇒ 2c³ - 3c² - 3c + 2 = 0

⇒ (c - 2)(2c² + c - 1) = 0

⇒ (c - 2) (2c² + 2c - c - 1) = 0

⇒ [(c - 2){2c(c + 1) - 1(c + 1)}] = 0

⇒ (c - 2)(2c - 1)(c + 1) = 0

$\tt{\rightarrow c=2\;,\;\dfrac{1}{2}\;,\; -1}$

Now,

$\textbf{\underline{c = -1\;is\;not\;possible}}$

$\textbf{\underline{Therefore,}}$

c = 2

⇒ t⁴(c)^⁶ = 64

⇒ t⁴(2)^⁶ = 64

⇒ t⁴ = 1

⇒ t = ± 1

$\textbf{\underline{The\;required\;numbers\;are :- }}$

${\boxed{\sf\:{1 , 2 , 4 , 8\;and\; -1 , -2 , -4 , - 8 }}}$

Substituting

$\tt{\rightarrow c=\dfrac{1}{2}}$

⇒t⁴(c)^⁶ = 64

$\tt{\rightarrow t^4(\dfrac{1}{2})^{6}=64}$

⇒ t⁴ = 64 × 64

⇒ t⁴ = 4096

⇒ t = ±8

$\textbf{\underline{The\;required\;numbers\;are :- }}$

${\boxed{\sf\:{8 , 4 , 2 , 1\;and\; -8 - 4 , -2 , -1}}}$