Key Concepts and Formulas

• Arithmetic Progression (AP): A sequence where the difference between consecutive terms is constant (common difference, $d$).
  • $k$-th term: $a_k = a_1 + (k-1)d$
  • Sum of $m$ terms: $S_m = \frac{m}{2}(a_1 + a_m)$ or $S_m = \frac{m}{2}(2a_1 + (m-1)d)$
• Sub-sequence from an AP: If we select terms from an AP at a constant interval, these selected terms also form an AP. If the original common difference is $d$ and we pick every $k$-th term, the new common difference is $kd$.

Step-by-Step Solution

Step 1: Understand the Given Information and Define Variables We are given an arithmetic progression $\{a_i\}_{i=1}^n$ with common difference $d=1$. We are also given two sums:

1. The sum of all $n$ terms: $\sum_{i=1}^n a_i = 192$.
2. The sum of the even-indexed terms: $\sum_{i=1}^{n/2} a_{2i} = 120$. We need to find the value of $n$, which is stated to be an even integer.

Step 2: Apply the Sum Formula to the First Condition The sum of the first $n$ terms of an AP is given by $S_n = \frac{n}{2}(a_1 + a_n)$. Using the first condition: $\sum_{i=1}^n a_i = \frac{n}{2}(a_1 + a_n) = 192$ From this, we can express the sum of the first and last terms: $a_1 + a_n = \frac{192 \times 2}{n} = \frac{384}{n} \quad \text{(Equation 1)}$

Step 3: Analyze and Apply the Sum Formula to the Second Condition The second condition is the sum of even-indexed terms: $a_2 + a_4 + a_6 + \dots + a_n = 120$. These terms form an AP. Let's determine its properties:

• First term: The first term of this new AP is $a_2$. Since $d=1$, $a_2 = a_1 + (2-1)d = a_1 + 1$.
• Common difference: The terms are $a_2, a_4, a_6, \dots$. The difference between consecutive terms is $a_4 - a_2 = (a_1 + 3d) - (a_1 + d) = 2d$. Since $d=1$, the common difference of this new AP is $D = 2 \times 1 = 2$.
• Number of terms: The sum is given as $\sum_{i=1}^{n/2} a_{2i}$. This means there are $n/2$ terms in this AP.
• Last term: The last term of this AP is $a_n$.

Now, we apply the sum formula for this AP. The sum of $m$ terms is $\frac{m}{2}(\text{first term} + \text{last term})$. Here, $m = n/2$, the first term is $a_2$, and the last term is $a_n$. $\frac{n/2}{2}(a_2 + a_n) = 120$ $\frac{n}{4}(a_2 + a_n) = 120$ Substitute $a_2 = a_1 + 1$: $\frac{n}{4}((a_1 + 1) + a_n) = 120$ $\frac{n}{4}(a_1 + a_n + 1) = 120$ Rearranging to isolate the term in the parenthesis: $a_1 + a_n + 1 = \frac{120 \times 4}{n} = \frac{480}{n} \quad \text{(Equation 2)}$

Step 4: Solve the System of Equations for n We have two equations:

1. $a_1 + a_n = \frac{384}{n}$
2. $a_1 + a_n + 1 = \frac{480}{n}$

Substitute the expression for $(a_1 + a_n)$ from Equation 1 into Equation 2: $\left(\frac{384}{n}\right) + 1 = \frac{480}{n}$ To solve for $n$, we can subtract $\frac{384}{n}$ from both sides: $1 = \frac{480}{n} - \frac{384}{n}$ Since the denominators are the same, we can combine the numerators: $1 = \frac{480 - 384}{n}$ $1 = \frac{96}{n}$ Multiplying both sides by $n$: $n = 96$

Common Mistakes & Tips

• Common Difference of Even Terms: A common error is to assume the common difference of the even-indexed terms ($a_2, a_4, \dots$) is still $d=1$. Remember that taking every second term of an AP with common difference $d$ results in a new AP with common difference $2d$.
• Number of Terms: Carefully identify the number of terms in each sum. The sum $\sum_{i=1}^{n/2} a_{2i}$ has $n/2$ terms, not $n$.
• Algebraic Manipulation: Ensure accuracy when combining fractions or rearranging equations to avoid arithmetic errors.

Summary The problem involves an arithmetic progression where we are given the sum of all terms and the sum of the even-indexed terms. By applying the formula for the sum of an arithmetic progression to both conditions, we derived two equations involving $a_1$, $a_n$, and $n$. Recognizing that the even-indexed terms also form an arithmetic progression with a common difference of $2d$ was crucial. Substituting the expression for the sum of the first and last terms from the first condition into the second condition allowed us to form an equation solely in terms of $n$, which we then solved to find $n=96$.

The final answer is \boxed{96}.