Question

a₁, a₂, a₃, … aₙ is a geometric sequence
follow a₂ = \\(\frac{4}{25}\\) a₄ = \\(\frac{16}{625}\\) r > 0
find \\(\sum_{n=1}^\infty a_n\\)

Explanation:

Step1: Find common ratio $r$

For geometric sequence, $a_4 = a_2 r^2$. So $r^2 = \frac{a_4}{a_2} = \frac{16/625}{4/25} = \frac{4}{25}$. Since $r>0$, $r = \frac{2}{5}$.

Step2: Find first term $a_1$

$a_2 = a_1 r \implies a_1 = \frac{a_2}{r} = \frac{4/25}{2/5} = \frac{2}{5}$.

Step3: Sum infinite geometric series

Sum formula: $S = \frac{a_1}{1 - r} = \frac{2/5}{1 - 2/5} = \frac{2/5}{3/5} = \frac{2}{3}$? Wait no, wait: Wait $a_2 = 4/25 = a_1*(2/5) \implies a_1 = (4/25)/(2/5) = (4/25)*(5/2) = 2/5$. Then sum is $a_1/(1 - r) = (2/5)/(1 - 2/5) = (2/5)/(3/5) = 2/3$. Wait wait no, wait let's recheck: $a_2 = 4/25$, $a_4 = a_2*r^2 = 16/625$. So $r^2 = (16/625)/(4/25) = (16/625)*(25/4) = 4/25$, so $r=2/5$. Correct. Then $a_1 = a_2 / r = (4/25)/(2/5) = 2/5$. Correct. Then sum is $a_1/(1 - r) = (2/5)/(3/5) = 2/3$. Yes, that's right.

Wait wait I made a mistake earlier in the answer line. Let's correct:

Step1: Calculate common ratio $r$

$r^2 = \frac{a_4}{a_2} = \frac{16/625}{4/25} = \frac{4}{25} \implies r = \frac{2}{5}$ (since $r>0$).

Step2: Find first term $a_1$

$a_1 = \frac{a_2}{r} = \frac{4/25}{2/5} = \frac{2}{5}$.

Step3: Compute infinite sum

$S = \frac{a_1}{1 - r} = \frac{2/5}{1 - 2/5} = \frac{2}{3}$.

Answer:

$\frac{4}{5}$