Animated Solution for Mathematics - Sequence and Series: Let $x_1,x_2,\dots ,x_{100}$ be in an arithmetic progression, with $x_1=2$ and their mean equal to 200. If $y_i=i(x_i-i),1\le i\le 100$, then the mean of $y_1,y_2,\dots ,y_{100}$ is

A Journey into Arithmetic Progressions

Welcome, future engineer. Today, we are not just solving a problem; we are peeling back the layers of a sequence to reveal the elegant structure hidden beneath. In the world of JEE Advanced, problems like this are not tests of your ability to calculate; they are tests of your ability to see the underlying geometry of numbers.

Let us embark on this journey together.

The Detective Work

We are given a sequence $x_1,x_2,\dots ,x_{100}$ that forms an Arithmetic Progression (A.P.). We know the first term, $x_1=2$, and we are given the mean of all one hundred terms: $200$.

Most students immediately reach for the sum formula, $S_n=\frac{n}{2}(2a+(n-1)d)$. While correct, it is often the long way around. Instead, let us use the symmetry of an A.P.

The mean of an A.P. is simply the average of the first and last terms. If the mean is $200$, then:

Substituting $x_1=2$, we get $\frac{2+x_{100}}{2}=200$, which implies $2+x_{100}=400$. Thus, $x_{100}=398$.

See how quickly the fog clears? We have identified the boundaries of our sequence. Now, to define the sequence completely, we need the common difference, $d$.

Using the general term formula $x_n=x_1+(n-1)d$, we substitute our known values:

With $d=4$, we have the DNA of our sequence. The general term $x_i$ is simply $x_1+(i-1)d$, which simplifies to $2+(i-1)4=4i-2$. This is the key that unlocks the entire problem.

The Transformation

Now, the problem introduces a new sequence, $y_i$, defined as $y_i=i(x_i-i)$. This is where many students panic. They see a new variable and assume the complexity has doubled.

But look closer. We already know $x_i=4i-2$. Let us substitute this into the definition of $y_i$:

Suddenly, the terrifying $y_i$ has transformed into a simple quadratic expression. We are no longer dealing with an abstract sequence; we are dealing with a sum of squares and a sum of integers. This is the power of algebraic substitution—it turns the unknown into the familiar.

The Power of Summation

Our goal is to find the mean of $y_i$. The mean $\bar{y}$ is defined as the sum of all $y_i$ divided by the number of terms, $100$:

Here, we invoke the linearity of summation. We can distribute the summation sign across the terms and pull out the constants:

Now, we rely on our toolkit of standard summation formulas. These are the bread and butter of JEE mathematics:

1. The sum of the first $n$ squares: ${\sum }_{i=1}^n{i}^2=\frac{n(n+1)(2n+1)}{6}$ 2. The sum of the first $n$ integers: ${\sum }_{i=1}^{n}i=\frac{n(n+1)}{2}$

Substituting $n=100$ into these formulas, we get:

Notice the beauty of the cancellation. The $100$ outside the bracket will cancel with the $100$ in the numerators of both terms inside. Let us simplify:

Simplifying the fractions:

And there it is. We started with a sequence, transformed it, and used the fundamental laws of summation to arrive at the answer.

Remember, in JEE Advanced, the complexity is often a facade. By finding the general term $x_i$ first, you strip away the mystery. By using the linearity of summation, you break the problem into manageable pieces.

And by keeping your cool during the arithmetic, you ensure that your logic leads to the correct destination. You have the tools; now, trust them. Keep practicing, keep questioning, and keep falling in love with the process.