Question

Find four numbers.In G.P such that their product is 64 and sum of 2nd and 3rd number is 6​

Answer 1

Answer:

1,2,4,8

Step-by-step explanation:

Let the numbers be a,b,c,d

By problem ,

b+ c= 6

=> c = 6 - b

SINCE the number are in GP,

so , we can say that ,

=> b ² = ac

=> a = (b²)/c

Also , c² = bd ,

=> d = (c²)/b

BY PROBLEM,

=> a*b*c*d = 64

=> {(b²)/c}*b*c*{(c²)/b} =  64

=> b²c²=64

=> bc = √64 = 8

=> b(6-b)=8

=> 6b - b² = 8

=> b² - 6b +8 =0

=> (b-4)(b-2)=0

=> b= 2

so, c =6-2 = 4

=> a = b²/c = 4/4 = 1

=> d = c²/b= 14/2 = 8

so , the numbers are = 1,2,4,8

Answer 2

Step-by-step explanation:

Given:-

Find four numbers in G. P. such that their product is 64 and sum of the second and third number is 6.

To find:-

The four number in G.P.

Solution:-

$\tt{Let \: four \: number \: be \: \: \frac{a}{ {r}^{3} }, \frac{a}{r} ,ar, {ar}^{3} }$

$\tt{(Common \: ratio \: is \: {r}^{2} }$

According to the first condition

$\tt{ \frac{a}{ {r}^{3} } \times \frac{a}{r} \times ar \times {ar}^{3} = 64}$

$∴ \: \: {a}^{4} = 64$

$∴ \: \: a = ( {2}^{6} )^{ \frac{1}{4} } = {2}^{ \frac{3}{2} }$

$∴ \: \: a = 2 \sqrt{2} .$

$\tt{Now \: using \: second \: condition \: \: \frac{a}{r} + ar = 6 }$

$\frac{2 \sqrt{2} }{r} + 2 \sqrt{2} \: r = 6.$

$\tt{Multiplying \: by \: \: r,}$

$\tt{2 \sqrt{2} + 2 \sqrt{2} \: {r}^{2} = 6r}$

$\tt{Dividing \: by \: 2}$

$\tt{\sqrt{2} + \sqrt{2} \: {r}^{2} = 3r}$

$\sqrt{2} \: {r}^{2} - 3r + \sqrt{2} = 0,$

$\tt{ \sqrt{2} \: {r}^{2} - 2r - r + \sqrt{2} = 0,}$

$\tt{ \sqrt{2} \: r \: (r - \sqrt{2} ) - 1(r - \sqrt{2} ) = 0.}$

$\tt{(r - \sqrt{2} ) \: ( \sqrt{2} \: r - 1) = 0}.$

$\tt{r = \sqrt{2} \: \: or \: \: r = \frac{1}{ \sqrt{2} }}$

If a = 2√2 , r = √2 then 1, 2, 4, 8 are the four required numbers in G.P.

If a = 2√2 , r = 1/√2 then 8, 4, 2, 1 are the four required numbers in G.P.

I hope it's help you..☺