Question 1209805
Let's solve this problem step-by-step.

Understanding Geometric Series

A geometric series has the form: a, ar, ar², ar³, ...
a is the first term (a_1)
r is the common ratio
a_n = ar^(n-1)
Given Information

The series has positive terms, so a > 0 and r > 0.
a_2 = 10
Finding a_1 and r

a_2 = ar^(2-1) = ar
ar = 10
a = 10/r
Finding a_1 + a_2 + a_3

a_1 + a_2 + a_3 = a + ar + ar²
Substitute a = 10/r:
(10/r) + 10 + 10r
Minimizing a_1 + a_2 + a_3

Let S = (10/r) + 10 + 10r. We want to minimize S.

Take the derivative with respect to r:

dS/dr = -10/r² + 10
Set the derivative to zero and solve for r:

-10/r² + 10 = 0
10 = 10/r²
r² = 1
r = ±1
Consider positive r:

Since the terms are positive, r must be positive. Therefore, r = 1.
Check the second derivative:

d²S/dr² = 20/r³
When r = 1, d²S/dr² = 20 > 0, which means we have a minimum.
Find a_1:

a = 10/r = 10/1 = 10
Find a_1 + a_2 + a_3:

a_1 + a_2 + a_3 = 10 + 10 + 10 = 30
However, we need to consider the behavior of the function as r approaches 0 or infinity.

As r approaches 0, 10/r approaches infinity, so S approaches infinity.
As r approaches infinity, 10r approaches infinity, so S approaches infinity.
Therefore, the minimum value of a_1 + a_2 + a_3 occurs when r = 1, and the minimum value is 30.

Final Answer: The smallest possible value of a_1 + a_2 + a_3 is 30.