To solve this, let the first term of the infinite geometric series be $a$, and the common ratio be $r$. Since it is an infinite geometric series with $|r| < 1$, the sum of the infinite series is given by:

$S_{\infty} = \frac{a}{1 - r}.$

Step 1: Use the sum of the infinite series

We are told that the sum of the infinite series is 16:

\frac{a}{1 - r} = 16. \tag{1}

Step 2: Use the sum of the first 4 terms

The sum of the first $n$ terms of a geometric series is given by:

$S_n = a \frac{1 - r^n}{1 - r}.$

For $n = 4$, the sum is 15:

a \frac{1 - r^4}{1 - r} = 15. \tag{2}

Step 3: Solve the system of equations

We now have two equations:

1. $\frac{a}{1 - r} = 16$,
2. $\frac{a (1 - r^4)}{1 - r} = 15$.

From Equation (1), solve for $a$ in terms of $r$:

a = 16(1 - r). \tag{3}

Substitute Equation (3) into Equation (2):

$\frac{16(1 - r)(1 - r^4)}{1 - r} = 15.$

Cancel $1 - r$ (since $r \neq 1$):

$16(1 - r^4) = 15.$

Simplify:

$16 - 16r^4 = 15.$

$16r^4 = 1.$

$r^4 = \frac{1}{16}.$

Take the fourth root:

$r = \pm \frac{1}{2}.$

Step 4: Determine the sign of $r$

Since $|r| < 1$ and the terms are positive (as the series sum is positive), we take:

$r = \frac{1}{2}.$

Step 5: Find $a$

Substitute $r = \frac{1}{2}$ into Equation (3):

$a = 16(1 - \frac{1}{2}).$

$a = 16 \cdot \frac{1}{2}.$

$a = 8.$

Final Answer:

• The first term $a$ is $8$,
• The common ratio $r$ is $\frac{1}{2}$.

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Let me know if you'd like further clarification or a breakdown of any step.

Related Questions:

1. How would the series change if $r = -\frac{1}{2}$?
2. What is the sum of the first 10 terms of this series?
3. What is the ratio of the 3rd term to the 1st term in this series?
4. Can this geometric series converge if $r \geq 1$?
5. How does the sum formula for an infinite series change for different values of $r$?

Tip: Always verify the conditions for convergence ($|r| < 1$) before applying the infinite series formula!