Animated Solution for Mathematics - Sequence and Series: The $20^{th}$ term from the end of the progression $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\dots ,-129\frac{1}{4}$ is :-

Imagine you are standing on a vast, infinite number line. You are tasked with walking along a specific path—an Arithmetic Progression (AP).

You start at $20$, and with every step, you move to the left by a fixed amount, landing on $19\frac{1}{4}$, then $18\frac{1}{2}$, and so on, until you reach a final, distant destination at $-129\frac{1}{4}$.

The question asks you to find the $20^{th}$ term from the end. If you try to count backward from that final, negative destination, you are essentially walking against the flow of the sequence, which is a recipe for confusion. But what if you could just turn around?

Before we perform any magic, we must prepare our tools. In mathematics, as in life, clarity is the first step to success.

Our sequence is given as $20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\dots ,-129\frac{1}{4}$. Those mixed fractions are clunky. Let us convert them into improper fractions to make our algebra sing.

Our first term, $a_1$, is $20$. Our second term, $a_2$, is $\frac{77}{4}$.

The common difference, $d$, is the heartbeat of this progression. We calculate it as:

With a common denominator, this becomes $\frac{77-80}{4}$, giving us $d=-\frac{3}{4}$. The negative sign confirms our intuition: we are indeed marching toward the left, toward the negative numbers.

The Reversal Trick

Now, we face the challenge: finding the $20^{th}$ term from the end. Instead of counting backward, let us pick up this entire number line and flip it.

By reversing the sequence, the last term, $l=-129\frac{1}{4}$, becomes our new starting point, $a^{\prime }$. Converting this to an improper fraction, we get $a^{\prime }=-\frac{517}{4}$.

When we reverse the sequence, we are now walking in the opposite direction. If we were moving left by $\frac{3}{4}$ at each step, we are now moving right by $\frac{3}{4}$ at each step.

Thus, our new common difference, $d^{\prime }$, is simply the negative of our old $d$. So, $d^{\prime }=-(-\frac{3}{4})=\frac{3}{4}$. We have transformed a confusing 'counting from the end' problem into a straightforward 'find the $n^{th}$ term' problem.

With our new parameters, $a^{\prime }=-\frac{517}{4}$ and $d^{\prime }=\frac{3}{4}$, we use the general term formula for an AP: $T_n=a+(n-1)d$. We want the $20^{th}$ term, so $n=20$.

Substituting our values, we get:

This simplifies to:

Performing the multiplication, $19\times 3=57$, so we have:

Since the denominators are identical, we combine the numerators:

Finally, dividing $-460$ by $4$ gives us $-115$.

The journey is complete. We have navigated the sequence, reversed our perspective, and arrived precisely at our destination. The $20^{th}$ term from the end is $-115$.