Animated Solution for Mathematics - Sequence and Series: Let $\frac{1}{x_1},\frac{1}{x_2},\dots ,\frac{1}{x_n}$ ($x_i\ne 0$ for $i=1,2,\dots ,n$) be in A.P. such that $x_1=4$ and $x_{21}=20$. If n is the least positive integer for which $x_n>50$, then ${\sum }_{i=1}^n\left(\frac{1}{x_i}\right)$ is equal to :

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a detective mission. In the world of JEE Advanced, the most dangerous traps are not the ones that look difficult, but the ones that look simple.

We are given a sequence where $\frac{1}{x_1},\frac{1}{x_2},\dots ,\frac{1}{x_n}$ form an Arithmetic Progression (A.P.). Many students rush in, see $x_1=4$ and $x_{21}=20$, and immediately try to force an A.P. on $x_i$.

Stop! Breathe. The sequence of reciprocals is the one that follows the linear rhythm of an A.P. Let us define our variables clearly. Let the first term of this A.P. be $a=\frac{1}{x_1}$ and the common difference be $d$. Since $x_1=4$, our first term is $a=\frac{1}{4}$.

Finding the Rhythm

Every A.P. is defined by its first term and its common difference. We have the first term, but we are missing the common difference $d$. We are given $x_{21}=20$, which means the $21^{st}$ term of our reciprocal sequence is $\frac{1}{x_{21}}=\frac{1}{20}$.

Using the general term formula for an A.P., $T_k=a+(k-1)d$, we can write:

This simplifies to $\frac{1}{20}=\frac{1}{4}+20d$. Now, let us isolate $d$. Subtracting $\frac{1}{4}$ from both sides, we get:

Finding a common denominator of $20$, we have $20d=\frac{1-5}{20}=-\frac{4}{20}=-\frac{1}{5}$. Dividing by $20$, we find:

Notice the negative sign? It tells us that our sequence is decreasing. The reciprocals are getting smaller, which means the original terms $x_i$ are growing.

Crossing the Threshold

Now, we need to find the $n$-th term. The general expression for our reciprocal sequence is $\frac{1}{x_n}=a+(n-1)d$. Substituting our values, we get:

Therefore, $x_n=\frac{100}{26-n}$. The problem demands that $x_n>50$. So, we set up the inequality:

Since $x_n>50$ is positive, $26-n$ must be positive. We can safely cross-multiply: $100>50(26-n)$, which simplifies to $2>26-n$. Rearranging, we get $n>24$. The least positive integer satisfying this is $n=25$.

Summing the Series

We are asked to find the sum ${\sum }_{i=1}^n\left(\frac{1}{x_i}\right)$ for $n=25$. This is the sum of the first $25$ terms of our A.P. The formula is $S_n=\frac{n}{2}[2a+(n-1)d]$.

Substituting $n=25$, $a=\frac{1}{4}$, and $d=-\frac{1}{100}$:

Simplifying inside the brackets:

Converting $\frac{1}{2}$ to $\frac{50}{100}$, we get:

Canceling terms, $\frac{25}{100}$ becomes $\frac{1}{4}$, so $S_{25}=\frac{1}{2}\times \frac{26}{4}=\frac{13}{4}$.

The final result is $\frac{13}{4}$.