Animated Solution for Mathematics - Sequence and Series: If $\{a_i{\}}_{i=1}^n$, where $n$ is an even integer, is an arithmetic progression with common difference 1, and ${\sum }_{i=1}^n{a}_i=192$, ${\sum }_{i=1}^{n/2}{a}_{2i}=120$, then $n$ is equal to:

Unlocking the Hidden Pattern

Welcome, my dear student. Today, we are not just solving a problem; we are peeling back the layers of an Arithmetic Progression (AP) to reveal the elegant structure hidden beneath. Many students look at a sequence and see a list of numbers; I want you to look at this sequence and see a story—a story of growth, of patterns, and of mathematical harmony.

The Architecture of the Sequence

Imagine a long line of students standing at equal distances. This is the physical embodiment of an Arithmetic Progression. We have a sequence $\{a_i{\}}_{i=1}^n$ with a common difference $d=1$. We are told that $n$ is an even integer and the total sum of these terms is $192$.

We start with our master key: the sum formula for an AP:

This formula is the heartbeat of the sequence. It connects the total sum, the number of terms, the first term, and the common difference. Let us substitute our known values: $d=1$ and $S_n=192$.

Let us simplify this. Multiplying by $1$ changes nothing, so we have $2a_1+n-1$. By multiplying both sides by $2$ and dividing by $n$, we get:

This is our first pillar. It is a beautiful, simplified relationship between our first term $a_1$ and the total number of terms $n$. Hold onto this; it is the foundation of our solution.

The Mystery of the Sub-sequence

Now, the problem introduces a twist. We are given the sum of the even-indexed terms: $a_2,a_4,a_6,\dots ,a_n$. This sum is $120$.

Do not panic. This is not a new problem; it is a sub-problem. Let us analyze this new sub-sequence:

1. The First Term: The first term of this sub-sequence is $a_2$. Since $a_2=a_1+d$ and $d=1$, our new first term is $a_1+1$.

2. The Common Difference: The gap between $a_2$ and $a_4$ is $(a_1+3d)-(a_1+d)=2d$. Since $d=1$, our new common difference is $2$.

3. The Number of Terms: Since we are taking every second term from a total of $n$ terms, we have exactly $m=\frac{n}{2}$ terms.

The Second Pillar

Let us apply the sum formula again, but this time to our sub-sequence. The sum is $120$.

Substituting our new parameters ($m=n/2$, $a_1^{\prime }=a_1+1$, $d^{\prime }=2$):

Let us simplify this expression. The term outside becomes $\frac{n}{4}$. Inside the bracket, we expand:

Look at that! The $+2$ and $-2$ cancel out perfectly. We are left with:

Multiplying both sides by $\frac{4}{n}$, we arrive at our second pillar:

The Grand Finale

We now have two equations:

1) $2a_1+n-1=\frac{384}{n}$

2) $2a_1+n=\frac{480}{n}$

Look closely at these two equations. Both contain the expression $2a_1+n$. We simply subtract Equation (1) from Equation (2):

The $2a_1$ and $n$ terms vanish into thin air, leaving us with:

And just like that, the answer reveals itself. Multiplying by $n$, we get:

My dear student, look at what we have achieved. We did not brute-force our way through the algebra. We identified the structure, we defined the sub-sequence, and we used the beauty of elimination to bypass the tedious work. This is the mindset of a JEE topper. Keep this clarity, keep this curiosity, and you will conquer any problem that comes your way.