Question

If the geometric mean between k-2 and k+4 is 4,find the value of k?​

Answer 1

Answer:

if k−1 is the G.M between k−2 and k+1, then find the value of k

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Answer

G.M of (k−2)and(k+1) is (k−1)

(k−1)

2

=(k−2)(k+1)

k

2

−2k+1=k

2

−k−2

k=3

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Answer 2

$\purple{ \bf{ \: \large\underline{\sf{Given- }}}}$

The geometric mean between k-2 and k+4 is 4.

$\purple{ \bf{ \: \large\underline{\sf{To\:Find - }}}}$

The value of k.

$\red{\boxed{ \rm{ \: \large\underline{\sf{Solution-}}}}}$

Given that,

The geometric mean between k-2 and k+4 is 4.

We know,

Geometric mean, (g) between two real numbers a and b is given by

$\red{ \: \: \: \boxed{ \: \: \rm{ \: g = \sqrt{ab} \: \: \: }}}$

Here,

$\rm :\longmapsto\:a \: = \: k - 2$

$\rm :\longmapsto\:b \: = \: k + 4$

$\rm :\longmapsto\:g = 4$

So, on substituting the values, we get

$\rm :\longmapsto\:4 = \sqrt{(k - 2)(k + 4)}$

On squaring both sides, we get

$\rm :\longmapsto\:16 = (k - 2)(k + 4)$

$\rm :\longmapsto\: {k}^{2} + 4k - 2k - 8 = 16$

$\rm :\longmapsto\: {k}^{2} + 2k - 8 - 16 = 0$

$\rm :\longmapsto\: {k}^{2} + 2k -24 = 0$

$\rm :\longmapsto\: {k}^{2} + 6k - 4k -24 = 0$

$\rm :\longmapsto\:(k + 6)(k - 4) = 0$

$\bf\implies \:k = - \: 6 \: \: \: or \: \: \: k = 4$

Verification :-

When k = 4

Numbers are

$\rm :\longmapsto\:a \: = \: k - 2 = 4 - 2 = 2$

$\rm :\longmapsto\:b \: = \: k + 4 = 4 + 4= 8$

$\bf\implies \:g = \sqrt{(2)(8)} = \sqrt{16} = 4$

Hence, Verified

Case :- 2

When k = - 6

Numbers are

$\rm :\longmapsto\:a \: = \: k - 2 = - 6 - 2 = - 8$

$\rm :\longmapsto\:b \: = \: k + 4 = - 6 + 4= - 2$

$\bf\implies \:g = \sqrt{( - 2)( - 8)} = \sqrt{16} = 4$

Hence, Verified

More information :-

1. Arithmetic Mean between two numbers a and b is

$\red{\boxed{ \rm{ \: \: \: AM = \frac{a + b}{2} \: \: \: }}}$

2. Arithmetic mean > Geometric Mean > Harmonic mean

3. Harmonic mean between two numbers a and b is

$\red{\boxed{ \rm{ \: \: \: \: HM \: = \: \frac{2ab}{a + b} \: \: \: \: }}}$