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viva [34]
2 years ago
13

Show that b^2 is greater than equal to or less than ac, according as a, b, c are in A.P., G.P.

Mathematics
1 answer:
ValentinkaMS [17]2 years ago
5 0

Answer/Step-by-step explanation  (ac > b² or b² < ac. )

A/c to question, we have to show:-

b² >ac in A.P ........ (1)

b² = ac in G.P .....(2)

b² < ac in H.P. ..... (3)

b = a+c/2 (A.P)

b = √ac ( G.P)

b = 2ac/a+c (H.P)

In A.P :

b² > ac = b² - ac

= (a+c/2)² - ac

= (a²+2ac+c²/4) - ac  = a² + 2ac + c² - 4ac / 4

= a² - 2ac + c² / 4  = ( a - c ) ² / 4 > 0  Hence, b²>ac

In G.P:-

b = √ac

Hence, b² = ac

In H.P :-  b² < ac = ac > b²  = ac - b²  = ac - ( 2ac / a+c)

= ac(a+c) - 2ac / a+c

= a²c + ac² - 2ac / a+c

= ac(2ac - 2) / a+c > 0

Hence, ac > b² or b² < ac.

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If cos(x) = Three-fourths and tan(x) &lt; 0, what is cos(2x)?
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Step-by-step explanation:

The value of sin(2x) is \sin(2x) = - \frac{\sqrt{15}}{8}sin(2x)=−

8

15

How to determine the value of sin(2x)

The cosine ratio is given as:

\cos(x) = -\frac 14cos(x)=−

4

1

Calculate sine(x) using the following identity equation

\sin^2(x) + \cos^2(x) = 1sin

2

(x)+cos

2

(x)=1

So we have:

\sin^2(x) + (1/4)^2 = 1sin

2

(x)+(1/4)

2

=1

\sin^2(x) + 1/16= 1sin

2

(x)+1/16=1

Subtract 1/16 from both sides

\sin^2(x) = 15/16sin

2

(x)=15/16

Take the square root of both sides

\sin(x) = \pm \sqrt{15/16

Given that

tan(x) < 0

It means that:

sin(x) < 0

So, we have:

\sin(x) = -\sqrt{15/16

Simplify

\sin(x) = \sqrt{15}/4sin(x)=

15

/4

sin(2x) is then calculated as:

\sin(2x) = 2\sin(x)\cos(x)sin(2x)=2sin(x)cos(x)

So, we have:

\sin(2x) = -2 * \frac{\sqrt{15}}{4} * \frac 14sin(2x)=−2∗

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15

∗

4

1

This gives

\sin(2x) = - \frac{\sqrt{15}}{8}sin(2x)=−

8

15

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