ISI MS(QMBA) Chapter Practice

To prove that the sequence converges and find its limit, we will follow these steps: 1. Find the potential limit (fixed point). 2. Prove that the sequence is bounded (specifically, bounded above). 3. Prove that the sequence is monotonic (specifically, monotonically increasing). 4. Conclude convergence using the Monotonic Bounded Theorem. **Step 1: Find the potential limit.** If the sequence converges to a limit $L$, then as $n \to \infty$, $x_n \to L$ and $x_{n+1} \to L$. Substituting this into the recurrence relation: Square both sides: Factor the quadratic equation: This gives two potential limits: $L=2$ or $L=-1$. Since $x_1=1$ and all subsequent terms $x_{n+1} = \sqrt{2+x_n}$ must be positive (as the square root of a positive number), $L$ must be non-negative. Therefore, the only possible limit is $L=2$. **Step 2: Prove the sequence is bounded above.** We will prove by induction that $x_n < 2$ for all $n \ge 1$. * **Base Case (n=1):** $x_1 = 1$, which is clearly less than $2$. So, $x_1 < 2$ is true. * **Inductive Hypothesis:** Assume that $x_k < 2$ for some integer $k \ge 1$. * **Inductive Step:** We need to show that $x_{k+1} < 2$. From the inductive hypothesis, $x_k < 2$. Adding $2$ to both sides: $2 + x_k < 2 + 2 = 4$. Taking the square root of both sides (since all terms are positive): $\sqrt{2 + x_k} < \sqrt{4}$. By definition, $x_{k+1} = \sqrt{2 + x_k}$. So, $x_{k+1} < 2$. Thus, by mathematical induction, $x_n < 2$ for all $n \ge 1$. This means the sequence is bounded above by $2$. **Step 3: Prove the sequence is monotonically increasing.** We will prove by induction that $x_{n+1} > x_n$ for all $n \ge 1$. * **Base Case (n=1):** $x_1 = 1$. $x_2 = \sqrt{2 + x_1} = \sqrt{2+1} = \sqrt{3}$. Since $\sqrt{3} \approx 1.732$, we have $x_2 > x_1$ ($1.732 > 1$). So, $x_2 > x_1$ is true. * **Inductive Hypothesis:** Assume that $x_k > x_{k-1}$ for some integer $k \ge 2$. * **Inductive Step:** We need to show that $x_{k+1} > x_k$. From the inductive hypothesis, $x_k > x_{k-1}$. Adding $2$ to both sides: $2 + x_k > 2 + x_{k-1}$. Taking the square root of both sides: $\sqrt{2 + x_k} > \sqrt{2 + x_{k-1}}$. By definition, $x_{k+1} = \sqrt{2 + x_k}$ and $x_k = \sqrt{2 + x_{k-1}}$. So, $x_{k+1} > x_k$. Thus, by mathematical induction, $x_{n+1} > x_n$ for all $n \ge 1$. This means the sequence is monotonically increasing. **Step 4: Conclude convergence.** From Step 2, the sequence $\{x_n\}$ is bounded above by $2$. From Step 3, the sequence $\{x_n\}$ is monotonically increasing. According to the Monotonic Bounded Theorem, every monotonic and bounded sequence converges to a finite limit. Since we found the only possible positive limit to be $L=2$ in Step 1, and the sequence starts at $1$ and is increasing, it must converge to $2$. Therefore, the sequence converges to $2$. **Common Mistakes:** 1. **Assuming convergence:** Many students solve $L = \sqrt{2+L}$ to find the limit but fail to prove that the sequence actually converges (i.e., proving monotonicity and boundedness). Finding the fixed point is only a potential limit, not a guarantee of convergence. 2. **Incorrectly solving for L:** Algebraic errors in solving the quadratic equation $L^2 - L - 2 = 0$ or not considering which roots are valid in the context of the sequence (e.g., discarding $L=-1$ because sequence terms are positive). 3. **Errors in inductive proofs:** Mistakes in setting up the base case, inductive hypothesis, or inductive step for either monotonicity or boundedness. Forgetting to state the inductive principle explicitly. 4. **Misapplying the Monotonic Bounded Theorem:** Not clearly stating that the sequence needs to be BOTH monotonic AND bounded to apply the theorem. Answer: $\boxed{2}$

Let the Harmonic Progression be $a_1, a_2, \ldots, a_k, \ldots$. According to the definition of HP, the reciprocals of its terms form an Arithmetic Progression (AP). Let this AP be $T_1, T_2, \ldots, T_k, \ldots$, where $T_k = \frac{1}{a_k}$. Let the first term of this AP be $A$ and its common difference be $D$. The general term of an AP is $T_k = A + (k-1)D$. Given that the $m$-th term of the HP is $n$, so $a_m = n$. This means the $m$-th term of the corresponding AP is $T_m = \frac{1}{n}$. Thus, we have: Given that the $n$-th term of the HP is $m$, so $a_n = m$. This means the $n$-th term of the corresponding AP is $T_n = \frac{1}{m}$. Thus, we have: To find $D$, subtract equation (2) from equation (1): Since $m \neq n$, we can divide by $(m-n)$: Now, substitute the value of $D$ back into equation (1) to find $A$: We need to find the $(m+n)$-th term of the HP, which is $a_{m+n}$. This is the reciprocal of the $(m+n)$-th term of the AP, $T_{m+n}$. Substitute the values of $A$ and $D$: Finally, the $(m+n)$-th term of the HP is the reciprocal of $T_{m+n}$: Therefore, the $(m+n)$-th term of the HP is $\frac{mn}{m+n}$. Answer: $\boxed{\frac{mn}{m+n}}$ **Why option D is correct:** Option D, $\frac{mn}{m+n}$, is the correct result derived from applying the definition of HP and the formula for the $n$-th term of an AP. **Why other options are incorrect:** * **A. $m+n$**: This would be the $(m+n)$-th term if the sequence were a special AP where the $m$-th term is $n$ and $n$-th term is $m$ (with $A=m+n-1, D=-1$), or if there was a direct additive property that does not apply to HP. * **B. $mn$**: This result is often seen in certain AP/GP problems, but it doesn't directly follow from the given conditions for an HP. It could result from algebraic errors or confusion with product relationships. * **C. $1$**: This would imply that $mn = m+n$, which is only true if $m=2, n=2$ (but $m \neq n$) or for specific non-integer values, which is not generally true for arbitrary $m, n$. **Common Mistakes:** 1. **Confusing HP with AP/GP properties:** Directly applying AP or GP formulas to HP terms without first converting to the reciprocal AP, leading to incorrect calculations. 2. **Algebraic errors in solving simultaneous equations:** Mistakes in subtracting equations or substituting values for $A$ and $D$, especially when dealing with fractions. 3. **Forgetting to take the reciprocal:** Calculating the $(m+n)$-th term of the corresponding AP but forgetting that the final answer for the HP requires taking the reciprocal of that term. 4. **Assuming specific values for m and n:** While testing with specific values can help verify, it's crucial to derive the general formula as required for a proof or a general question, and specific values might not catch all algebraic nuances.

**Part i: Show that $f(n)$ forms a Geometric Progression for $n \in \mathbb{N}$, and state its first term and common ratio.** Step 1: Use the given functional relation $f(x+y) = f(x)f(y)$ and $f(1)=2$. For $n=1$: $f(1) = 2$ For $n=2$: $f(2) = f(1+1) = f(1)f(1) = 2 \cdot 2 = 2^2 = 4$. For $n=3$: $f(3) = f(2+1) = f(2)f(1) = 4 \cdot 2 = 2^3 = 8$. Step 2: Generalize the pattern. It appears that $f(n) = 2^n$. We can prove this by induction. Base case: For $n=1$, $f(1)=2^1=2$, which is true. Assume $f(k) = 2^k$ for some natural number $k$. Consider $f(k+1) = f(k+1-1+1) = f(k)f(1)$ (using the functional relation). Substitute the assumption $f(k)=2^k$ and the given $f(1)=2$: $f(k+1) = 2^k \cdot 2 = 2^{k+1}$. Thus, by induction, $f(n) = 2^n$ for all $n \in \mathbb{N}$. Step 3: Show $f(n)$ forms a GP and state its first term and common ratio. The sequence is $f(1), f(2), f(3), \dots$, which is $2^1, 2^2, 2^3, \dots$. The first term is $f(1) = 2$. The ratio of consecutive terms is $\frac{f(n+1)}{f(n)} = \frac{2^{n+1}}{2^n} = 2$. Since the ratio is a constant ($2$), the sequence $f(n)$ forms a Geometric Progression with first term $a=2$ and common ratio $r=2$. Answer: First term $\boxed{a=2}$, Common ratio $\boxed{r=2}$ **Part ii: Determine the value of $\sum_{m=1}^{n} f(m)$.** Step 1: Write out the sum using the derived form of $f(m)$. Step 2: Identify the parameters of this Geometric Progression. First term $A = 2$. Common ratio $R = 2$. Number of terms $N = n$. Step 3: Apply the formula for the sum of $n$ terms of a GP, $S_N = \frac{A(R^N - 1)}{R - 1}$. Answer: $\boxed{2(2^n - 1)}$ **Part iii: If the sum $\sum_{m=1}^{n} f(m)$ is expressed in the form $C(2^n - 1)$, find the value of the constant $C$.** Step 1: Equate the sum derived in Part ii with the given form. We found that $\sum_{m=1}^{n} f(m) = 2(2^n - 1)$. The problem states that the sum is $C(2^n - 1)$. Step 2: Solve for $C$. For this equality to hold for all $n \in \mathbb{N}$ (specifically for $n > 0$, so $2^n-1 \neq 0$), Answer: $\boxed{C=2}$ **Common Mistakes:** 1. **Incorrectly identifying the common ratio or first term:** Students might confuse $f(0)$ as the first term if they extend the domain to $x=0$, or make errors in calculating the ratio. Always explicitly state $f(1)$ as the first term for $n \in \mathbb{N}$. 2. **Errors in applying the sum formula:** A common mistake is using the incorrect formula for $S_n$ or making algebraic errors during substitution. 3. **Algebraic errors when solving for C:** Simple algebraic mistakes can lead to an incorrect value of $C$. Ensuring $2^n-1$ is cancelled correctly is important. 4. **Not providing a formal proof for $f(n)=2^n$:** For a subjective question asking to 'show', a clear inductive or step-by-step derivation for the general term is necessary, not just stating the pattern.

Step 1: Find the potential limits (fixed points) of the sequence. Assume the sequence converges to a limit $L$. Then, as $n \to \infty$, we have $\lim_{n \to \infty} a_{n+1} = L$ and $\lim_{n \to \infty} a_n = L$. Substituting these into the recurrence relation: Square both sides (note that since $a_n > 0$ for all $n$, $L$ must be non-negative): Factor the quadratic equation: This gives two potential limits: $L=3$ or $L=-1$. Since $a_1 = 1 > 0$, and $a_{n+1} = \sqrt{2a_n + 3}$ will always be positive if $a_n$ is positive (as $2a_n+3 > 0$), all terms of the sequence must be positive. Therefore, the limit $L$ must also be non-negative. This eliminates $L=-1$. So, the only possible limit is $L=3$. Step 2: Prove that the sequence is monotonically increasing. We want to show that $a_{n+1} > a_n$ for all $n$. This is equivalent to Since all terms $a_n$ are positive, we can square both sides without changing the inequality direction: Factor the quadratic expression: Since $a_n > 0$, $a_n+1$ is always positive. For the product to be negative, we must have $a_n - 3 < 0$, which means $a_n < 3$. So, the sequence is monotonically increasing if $a_n < 3$. Let's check the first few terms: Indeed, $a_1 < a_2 < a_3$, and all these terms are less than $3$. This suggests the sequence is increasing and bounded above by $3$. Step 3: Prove that the sequence is bounded above by $3$. We will use induction to prove that $a_n < 3$ for all $n \ge 1$. Base Case: For $n=1$, $a_1 = 1$, which is clearly less than $3$. So, $a_1 < 3$ is true. Inductive Step: Assume that $a_k < 3$ for some arbitrary positive integer $k$. We need to show that $a_{k+1} < 3$. From the recurrence relation, Since we assumed $a_k < 3$, we have: Add $3$ to both sides: Take the square root of both sides (since $2a_k+3 > 0$): Thus, by the principle of mathematical induction, $a_n < 3$ for all $n \ge 1$. The sequence is bounded above by $3$. Also, since $a_1=1$ and $a_{n+1} = \sqrt{2a_n+3}$, all terms are positive, so the sequence is bounded below by $1$ (or $0$). Step 4: Conclude convergence and find the limit. From Step 2, we showed that the sequence is monotonically increasing because $a_n < 3$ for all $n$. From Step 3, we showed that the sequence is bounded above by $3$. According to the Monotonic Bounded Theorem, any sequence that is both monotonic (increasing in this case) and bounded (above) must converge to a finite limit. From Step 1, the only feasible limit for this sequence is $L=3$. Therefore, the sequence converges to $3$. The final answer is $\boxed{\text{Converges to 3}}$ **Common Mistakes:** 1. **Only finding fixed points without proving convergence:** Many students stop after solving $L = f(L)$ and assume the sequence converges to one of the solutions. It is crucial to prove monotonicity and boundedness to guarantee convergence. 2. **Not verifying the feasibility of fixed points:** In this case, $L=-1$ was a mathematical solution but not a valid limit for a sequence whose terms are always positive. Always check if the fixed points are consistent with the nature of the sequence (e.g., positive terms, real values). 3. **Errors in algebraic manipulation for monotonicity:** Incorrectly solving inequalities or forgetting to consider the sign of terms when squaring or dividing in inequalities can lead to wrong conclusions about monotonicity. 4. **Flawed inductive proof for boundedness:** Mistakes in the base case or the inductive step can lead to an incorrect conclusion about the sequence's bounds, which is essential for applying the Monotonic Bounded Theorem.

Step 1: Identify the components of the Arithmetico-Geometric Progression (AGP). Let the given series be $S$. The terms in the numerators form an Arithmetic Progression (AP): $1, 4, 7, 10, \dots$ For this AP, the first term is $a_{AP} = 1$ and the common difference is $d_{AP} = 4-1 = 3$. The terms involving powers of $1/2$ form a Geometric Progression (GP) multiplier: $1, \frac{1}{2}, \left(\frac{1}{2}\right)^2, \left(\frac{1}{2}\right)^3, \dots$ For this GP, the common ratio is $r = \frac{1}{2}$. Since $|r| = \frac{1}{2} < 1$, the sum to infinity exists. Step 2: Multiply the series by the common ratio $r$. Multiply Equation (1) by $r = \frac{1}{2}$: Step 3: Subtract Equation (2) from Equation (1), shifting terms appropriately. Step 4: Identify the resulting series as a Geometric Progression. The terms $3\left(\frac{1}{2}\right) + 3\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right)^3 + \dots$ form an infinite GP. For this new GP, the first term is $A = 3\left(\frac{1}{2}\right) = \frac{3}{2}$ and the common ratio is $R = \frac{1}{2}$. The sum to infinity of this GP is $S'_{inf} = \frac{A}{1-R}$. Step 5: Substitute $S'_{inf}$ back into Equation (3) and solve for $S$. **Why this solution is correct:** The method of multiplying by the common ratio and subtracting is the standard and most robust way to find the sum of an Arithmetico-Geometric Progression. This method successfully transforms the AGP into a simpler sum involving a constant term and an infinite Geometric Progression, whose sum can be readily calculated. The condition $|r|<1$ ensures the convergence of both the original AGP and the derived GP, making the sum well-defined. **Common Mistakes:** 1. **Incorrectly identifying AP/GP components:** Students might misinterpret the first term or common difference/ratio, leading to errors in the subsequent steps. 2. **Errors in subtraction and shifting terms:** A common mistake is not aligning terms correctly during subtraction, especially when dealing with the first term or missing the pattern after subtraction. 3. **Arithmetic errors:** Simple calculation mistakes, especially with fractions, can lead to an incorrect final sum. 4. **Forgetting to check for convergence:** While less likely to affect the final answer for convergent series, it's crucial to acknowledge that the sum to infinity only exists if $|r|<1$. Forgetting this condition indicates a conceptual gap. Answer: $\boxed{8}$

Step 1: Find the potential limit (fixed point) of the sequence. Assume the sequence converges to a limit $L$. Then, as $n \to \infty$, we have $\lim_{n \to \infty} a_{n+1} = L$ and $\lim_{n \to \infty} a_n = L$. Substituting these into the recurrence relation: Square both sides to remove the square root: Rearrange into a quadratic equation: Factor the quadratic equation: This gives two potential limits: $L=3$ or $L=-2$. Since $a_1 = 1$ and $a_{n+1} = \sqrt{6+a_n}$, all terms $a_n$ must be positive (as the square root of a non-negative number is non-negative, and $a_1=1$). Therefore, a negative limit like $L=-2$ is not possible. The only plausible limit is $L=3$. Step 2: Prove that the sequence is monotonically increasing. We need to show that $a_{n+1} > a_n$ for all $n \ge 1$. First, let's calculate the first few terms: Since $a_2 = \sqrt{7} > 1 = a_1$, the sequence appears to be increasing for $n=1$. Let's prove this by induction: Base case: For $n=1$, $a_2 = \sqrt{7} > 1 = a_1$. The base case holds. Inductive Hypothesis: Assume $a_k < a_{k+1}$ for some integer $k \ge 1$. We want to show $a_{k+1} < a_{k+2}$. We know $a_{k+1} = \sqrt{6+a_k}$ and $a_{k+2} = \sqrt{6+a_{k+1}}$. From the inductive hypothesis, $a_k < a_{k+1}$. Adding $6$ to both sides of the inequality: $6+a_k < 6+a_{k+1}$. Since the square root function is strictly increasing for non-negative inputs, we can take the square root of both sides while preserving the inequality: This implies $a_{k+1} < a_{k+2}$. Thus, by induction, the sequence is monotonically increasing. Step 3: Prove that the sequence is bounded above. We need to show that $a_n < 3$ for all $n \ge 1$. Base case: For $n=1$, $a_1 = 1 < 3$. The base case holds. Inductive Hypothesis: Assume $a_k < 3$ for some integer $k \ge 1$. We want to show $a_{k+1} < 3$. We know $a_{k+1} = \sqrt{6+a_k}$. From the inductive hypothesis, $a_k < 3$. So, adding $6$ to both sides: Taking the square root of both sides (since all terms are positive): Thus, by induction, the sequence is bounded above by $3$. Since $a_1=1$ and the sequence is increasing, it is also bounded below by $1$. Step 4: Conclude convergence using the Monotonic Bounded Theorem. Since the sequence $\{a_n\}$ is both monotonically increasing (from Step 2) and bounded above (from Step 3), by the Monotonic Bounded Theorem, the sequence converges to a finite limit. From Step 1, the only possible limit that aligns with the sequence's properties (positive terms) is $3$. Therefore, the sequence converges to $3$. **Common Mistakes:** 1. **Forgetting to verify convergence:** Simply finding the fixed point $L=3$ is not enough; one must also prove that the sequence actually converges to this limit by demonstrating it is monotonic and bounded. 2. **Incorrectly assuming monotonicity or boundedness:** These properties must be rigorously proven, typically by mathematical induction, rather than just observing the first few terms. 3. **Ignoring the domain of the square root:** When solving $L^2 = 6+L$, neglecting to consider that $L$ must be non-negative because $L = \lim a_n$ and all $a_n \ge 0$ (due to the square root definition and $a_1=1$). This would lead to accepting $L=-2$ as a possible limit. 4. **Algebraic errors in solving for L:** Mistakes in solving the quadratic equation for the fixed points. Answer: $\boxed{3}$

**Step-by-step Proof:** 1. **Understand the Given Condition:** We are given that $a, b, c$ are distinct non-zero real numbers in Harmonic Progression (HP). By the definition of HP, this means their reciprocals are in Arithmetic Progression (AP). Thus, $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP. 2. **State the Goal:** We need to prove that the sequence $X = \frac{a}{b+c}, Y = \frac{b}{c+a}, Z = \frac{c}{a+b}$ is in HP. This is equivalent to proving that their reciprocals, $\frac{1}{X}, \frac{1}{Y}, \frac{1}{Z}$, are in AP. So, we need to show that $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in AP. 3. **Manipulate the Terms of the New Sequence:** Let's consider the terms whose AP property we need to prove: First term: $\frac{b+c}{a}$ Second term: $\frac{c+a}{b}$ Third term: $\frac{a+b}{c}$ Let $S = a+b+c$. We can express each term using $S$: So, the sequence we need to check for AP is $S\left(\frac{1}{a}\right) - 1, S\left(\frac{1}{b}\right) - 1, S\left(\frac{1}{c}\right) - 1$. 4. **Apply Properties of Arithmetic Progression:** We know from Step 1 that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP. A fundamental property of APs is that if $A_1, A_2, A_3$ are in AP, then: * $k A_1, k A_2, k A_3$ are also in AP (for any non-zero constant $k$). * $A_1 + K, A_2 + K, A_3 + K$ are also in AP (for any constant $K$). In our case, $S$ is a constant ($a,b,c$ are fixed numbers). Since $a,b,c$ are non-zero, $S$ might be zero only if some are positive and some are negative, but the problem usually assumes positive numbers for HP context. If $S=0$, the argument still holds if $a,b,c$ are non-zero. Let's assume $S \neq 0$ to avoid division by zero issues in terms like $S/a$. Since $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP, multiplying each term by the constant $S$ ensures that $S\left(\frac{1}{a}\right), S\left(\frac{1}{b}\right), S\left(\frac{1}{c}\right)$ are also in AP. Further, subtracting the constant $1$ from each term preserves the AP property. Thus, $S\left(\frac{1}{a}\right) - 1, S\left(\frac{1}{b}\right) - 1, S\left(\frac{1}{c}\right) - 1$ are in AP. 5. **Conclusion:** Since the reciprocals of the sequence $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ form an Arithmetic Progression, the sequence itself is in Harmonic Progression. **Common Mistakes:** 1. **Direct Algebraic Substitution:** Many students attempt to directly substitute the relation $b = \frac{2ac}{a+c}$ into the terms $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$. This leads to extremely complex algebraic expressions that are very difficult to simplify and prove the AP property. 2. **Incorrect AP Property Application:** Incorrectly thinking that if $x_1, x_2, x_3$ are in AP, then $\frac{1}{x_1}, \frac{1}{x_2}, \frac{1}{x_3}$ are in HP. While true, this is the definition of HP, and its inverse is not always straightforward to apply in such manipulation. 3. **Ignoring Distinctness/Non-zero:** While not critical for this specific proof, in other HP problems, conditions like distinctness or non-zero terms can be crucial, especially when manipulating fractions or considering edge cases where common differences might be zero. 4. **Assuming $S=0$ is problematic:** If $a+b+c=0$, then $S=0$. In this case, the terms become $-1, -1, -1$, which are trivially in AP. The argument holds even for $S=0$ provided $a,b,c$ are non-zero.

Step 1: Identify the Arithmetic Progression (AP) and Geometric Progression (GP) parts of the series. The given series is an Arithmetico-Geometric Progression (AGP). Let's look at the numerators and denominators separately. Numerators: $1, 4, 7, 10, \dots$ This is an Arithmetic Progression with first term $a_{AP} = 1$ and common difference $d = 4 - 1 = 3$. Denominators (considering the sequence of fractions): $1, \frac{1}{5}, \frac{1}{25}, \frac{1}{125}, \dots$ This is a Geometric Progression with first term $a_{GP} = 1$ and common ratio $r = \frac{1}{5}$. For the AGP sum formula, we use the first term of the AP part as $a$, the common difference of the AP part as $d$, and the common ratio of the GP part as $r$. So, we have $a=1$, $d=3$, and $r=\frac{1}{5}$. Step 2: Check the condition for the convergence of an infinite AGP. An infinite AGP converges if and only if the absolute value of the common ratio of its GP part is less than 1, i.e., $|r| < 1$. Here, $|r| = |\frac{1}{5}| = \frac{1}{5} < 1$. So, the series converges. Step 3: Apply the formula for the sum of an infinite AGP. The sum to infinity of an AGP is given by the formula: Substitute the values $a=1$, $d=3$, and $r=\frac{1}{5}$ into the formula: Step 4: Perform the calculations. Calculate the terms: $1 - \frac{1}{5} = \frac{5}{5} - \frac{1}{5} = \frac{4}{5}$ So, the first part is: $\frac{1}{\frac{4}{5}} = \frac{5}{4}$ The second part is: $\frac{3 \cdot \frac{1}{5}}{(\frac{4}{5})^2} = \frac{\frac{3}{5}}{\frac{16}{25}}$ To simplify this, multiply by the reciprocal of the denominator: $\frac{3}{5} \cdot \frac{25}{16} = \frac{3 \cdot 5}{16} = \frac{15}{16}$ Now, add the two parts: To add these fractions, find a common denominator, which is 16: Therefore, the sum to infinity of the series is $\frac{35}{16}$. Why other options are incorrect: * "$\frac{25}{16}$": This result occurs if one mistakenly uses $d=1$ (the common difference of the AP part) instead of $d=3$. With $a=1, d=1, r=1/5$, the sum would be $\frac{1}{1-1/5} + \frac{1 \cdot 1/5}{(1-1/5)^2} = \frac{5}{4} + \frac{1/5}{16/25} = \frac{5}{4} + \frac{5}{16} = \frac{20+5}{16} = \frac{25}{16}$. * "$\frac{15}{4}$": This could result from significant arithmetic errors, perhaps by incorrectly combining terms or miscalculating the second part of the sum. For example, if the second term was mistakenly simplified as $\frac{3}{5} / (4/5) = 3/4$, then $5/4 + 3/4 = 8/4 = 2$, not $15/4$. Or if the entire expression was evaluated as $3 \cdot 5/4 = 15/4$, which is not how the formula works. * "$\frac{20}{9}$": This value does not correspond to a common or simple arithmetic error in the AGP sum formula for the given parameters. **Common Mistakes:** 1. Incorrectly identifying the common difference ($d$) of the AP part. Students might assume $d=1$ if not careful, instead of $d=3$ (from $1, 4, 7, \dots$). 2. Arithmetic errors, especially in squaring the denominator $(1-r)^2$ and in simplifying the compound fraction $\frac{dr}{(1-r)^2}$. 3. Forgetting the condition for convergence ($|r|<1$) for infinite series, although for this specific problem, it converges. 4. Confusing the formula for AGP sum with that of a simple GP, or misremembering parts of the AGP formula. Answer: $\boxed{\frac{35}{16}}$