Key Concepts and Formulas

• Sum of an Infinite Geometric Progression (G.P.): An infinite G.P. with first term $a$ and common ratio $r$ has a finite sum $S$ if and only if $|r| < 1$. The sum is given by the formula $S = \frac{a}{1-r}$.
• Convergence Condition: For the sum of an infinite G.P. to exist and be finite, the absolute value of the common ratio must be strictly less than 1, i.e., $|r| < 1$, which is equivalent to $-1 < r < 1$.

---

Step-by-Step Solution

Step 1: Understand the Problem and Identify Given Information We are given an infinite Geometric Progression (G.P.).

• The first term of the G.P. is denoted by $b$. So, $a = b$.
• The sum of this infinite G.P. is given as 5. So, $S = 5$. We need to find the interval in which the first term $b$ lies.

Step 2: Apply the Formula for the Sum of an Infinite G.P. The formula for the sum of an infinite G.P. is $S = \frac{a}{1-r}$. Substituting the given values, we have: $5 = \frac{b}{1-r}$ Explanation: This equation establishes a fundamental relationship between the first term ($b$), the common ratio ($r$), and the given sum (5). This will be our primary equation to manipulate.

Step 3: Incorporate the Convergence Condition for Infinite G.P.s For an infinite G.P. to have a finite sum, the common ratio $r$ must satisfy the condition $|r| < 1$. This is equivalent to: $-1 < r < 1$ Explanation: This condition is crucial. If $|r| \ge 1$, the series would diverge, and its sum would not be a finite value like 5. Therefore, any solution for $b$ must be consistent with this condition on $r$.

Step 4: Analyze the Sign of the First Term ($b$) From the equation $5 = \frac{b}{1-r}$, we can infer the sign of $b$. Since $|r| < 1$, we know that $r < 1$, which implies $1-r > 0$. The denominator $(1-r)$ is positive. Given that the sum $S = 5$ is positive, and the denominator $(1-r)$ is positive, the numerator $b$ must also be positive for the fraction to be positive. $b > 0$ Explanation: This step is a quick logical deduction that helps eliminate potential answer choices. If $b$ were negative, and $1-r$ were positive, the sum would be negative, contradicting $S=5$.

Step 5: Express the Common Ratio ($r$) in Terms of the First Term ($b$) We need to use the convergence condition $(-1 < r < 1)$ to find the range of $b$. To do this, we must express $r$ in terms of $b$ using the sum equation from Step 2. Rearranging $5 = \frac{b}{1-r}$: $5(1-r) = b$ $1-r = \frac{b}{5}$ $r = 1 - \frac{b}{5}$ Explanation: By isolating $r$, we establish a direct link between $r$ and $b$. This allows us to substitute the expression for $r$ into the inequality for $r$, thereby obtaining an inequality solely in terms of $b$.

Step 6: Substitute the Expression for $r$ into the Convergence Inequality Now, we substitute the expression $r = 1 - \frac{b}{5}$ into the convergence condition $-1 < r < 1$: $-1 < 1 - \frac{b}{5} < 1$ Explanation: This is the pivotal step where the convergence condition is applied to the relationship between $b$ and $r$. This transforms the problem into solving a compound inequality for $b$.

Step 7: Solve the Compound Inequality for $b$ We solve the inequality $-1 < 1 - \frac{b}{5} < 1$ by performing operations on all three parts:

• Subtract 1 from all parts: $-1 - 1 < \left(1 - \frac{b}{5}\right) - 1 < 1 - 1$ $-2 < -\frac{b}{5} < 0$

• Multiply all parts by -5: When multiplying an inequality by a negative number, we must reverse the direction of the inequality signs. $(-2) \times (-5) > \left(-\frac{b}{5}\right) \times (-5) > (0) \times (-5)$ $10 > b > 0$

• Rewrite in standard ascending order: $0 < b < 10$ Explanation: Each algebraic manipulation aims to isolate $b$. The multiplication by -5 requires careful attention to reversing the inequality signs. This step yields the final range for $b$.

Step 8: Determine the Interval for $b$ The solution to the inequality is $0 < b < 10$. This means that $b$ must be strictly greater than 0 and strictly less than 10. Therefore, $b$ lies in the open interval $(0, 10)$.

---

Common Mistakes & Tips

• Forgetting the Convergence Condition: The most common error is to overlook the requirement $|r| < 1$. This condition is essential for the existence of a finite sum and for determining the range of $b$.
• Errors with Inequality Signs: Be extremely cautious when multiplying or dividing inequalities by negative numbers. Always remember to reverse the direction of the inequality signs.
• Initial Sign Analysis of $b$: Since the sum $S=5$ is positive and $1-r$ must be positive (due to $|r|<1$), the first term $b$ must also be positive. This can quickly eliminate options where $b$ is negative or zero.

Summary

To find the interval for the first term $b$ of an infinite G.P. with a sum of 5, we first used the formula for the sum of an infinite G.P., $S = \frac{a}{1-r}$, setting $a=b$ and $S=5$. Crucially, we incorporated the convergence condition $|r|<1$, which implies $-1 < r < 1$. By rearranging the sum formula to express $r$ in terms of $b$, we substituted this expression into the convergence inequality. Solving the resulting compound inequality for $b$, while correctly handling the reversal of inequality signs when multiplying by a negative number, led us to the interval $0 < b < 10$.

The final answer is $\boxed{(0, 10)}$.