Question

Let three positive numbers $a, b, c$ are in geometric progression, such that $a, b+8, c$ are in arithmetic progression and $a, b+8, c+64$ are in geometric progression. If the arithmetic mean of $a, b, c$ is $k$, then $\frac{3}{13}k$ is equal to

Answer 1

4

Answer 2

Solution:

Let the numbers in the geometric progression be

$$a,\;b=ar,\;c=ar^2.$$

1. Since $a,\;b+8,\;c$ are in arithmetic progression,

$$b+8=\frac{a+c}{2} \quad \Longrightarrow \quad ar+8=\frac{a+ar^2}{2}.$$

Multiplying by 2,

$$2ar+16 = a(1+r^2) \quad \Longrightarrow \quad a(1+r^2-2r)=16. \quad (1)$$

2. Since $a,\;b+8,\;c+64$ are in geometric progression,

$$(b+8)^2=a(c+64).$$

Substitute $b=ar$ and $c=ar^2$:

$$(ar+8)^2 = a(ar^2+64).$$

Expanding,

$$a^2r^2+16ar+64 = a^2r^2+64a.$$

Cancel $a^2r^2$ from both sides:

$$16ar+64 = 64a \quad \Longrightarrow \quad 16ar = 64a-64.$$

Dividing by 16 (and noting $a>0$),

$$ar = 4a-4 \quad \Longrightarrow \quad a(4-r)=4,$$ $$a = \frac{4}{4-r}. \quad (2)$$

3. Substitute the value of $a$ from (2) into (1):

$$\frac{4}{4-r}(1+r^2-2r)=16.$$

Multiply both sides by $4-r$:

$$4(1+r^2-2r)=16(4-r),$$ $$1+r^2-2r =4(4-r)=16-4r.$$

Rearrange:

$$r^2-2r+1 =16-4r,$$ $$r^2+2r-15=0.$$

Factorizing,

$$(r+5)(r-3)=0.$$

Since $r>0$, we choose $r=3$.

4. Now, find $a$ using (2):

$$a=\frac{4}{4-3} = 4.$$

Then,

$$b=ar=4\times 3=12,\quad c=ar^2=4\times 9=36.$$

The arithmetic mean $k$ is:

$$k=\frac{a+b+c}{3}=\frac{4+12+36}{3}=\frac{52}{3}.$$

Thus,

$$\frac{3}{13}k=\frac{3}{13}\times \frac{52}{3}=\frac{52}{13}=4.$$

Core Explanation Summary:

• Expressed $a,\, b,\, c$ in GP.
• Used AP condition on $a,\, b+8,\, c$ to get equation (1).
• Used GP condition on $a,\, b+8,\, c+64$ to get equation (2).
• Solved for $r$ and $a$, then computed $b,\, c$ and $k$.
• Calculated $\frac{3}{13}k=4$.