Animated Solution for Mathematics - Sequence and Series: Let $a_n$ be $n^{th}$ term of the series $5+8+14+23+35+50+\dots$ and $S_n={\sum }_{k=1}^n{a}_k$. Then $S_{30}-a_{40}$ is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a sequence that, at first glance, seems to defy the standard rules of Arithmetic or Geometric Progressions.

Look at the series: $5,8,14,23,35,50,\dots$. It is not an A.P., because the difference between $8$ and $5$ is $3$, while the difference between $14$ and $8$ is $6$. It is not a G.P. either.

Let us look at the differences between consecutive terms: $8-5=3$, $14-8=6$, $23-14=9$, $35-23=12$, and $50-35=15$. The differences are $3,6,9,12,15,\dots$.

These differences form a perfect Arithmetic Progression with a first term $a=3$ and a common difference $d=3$. We have identified a sequence of the second order, and we have the perfect tool to solve it: the Method of Differences.

A Mathematical Dance

To find the general term $a_n$, we employ a beautiful algebraic technique. We write the sum of the series up to $n$ terms as $S=a_1+a_2+\cdots +a_n$.

Then, we write the same sum again, but shifted by one position: $S=a_1+a_2+\cdots +a_{n-1}+a_n$. When we subtract the second equation from the first, the terms align in a way that creates a telescoping effect.

The left side becomes zero, and on the right, we are left with the first term $a_1$ plus the sum of all the differences. Specifically:

Remember, because we are looking for the $n^{th}$ term, we only sum the first $n-1$ differences. This is a critical detail where many students stumble, so keep it locked in your memory!

Now, let us do the heavy lifting. We know $a_1=5$. The differences form an A.P. where the $k^{th}$ difference is $3k$.

So, $a_n=5+{\sum }_{k=1}^{n-1}3k$. Using the sum formula for an A.P., we have:

Thus, $a_n=5+\frac{3n(n-1)}{2}$. Simplifying this, we get:

This is the engine of our problem. For $a_{40}$, we simply plug in $n=40$:

The final phase of our journey is to find $S_{30}$, the sum of the first $30$ terms. We apply the summation operator to our general term:

By the linearity of summation, this becomes:

We use our standard formulas: $\sum n^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum n=\frac{n(n+1)}{2}$. For $n=30$:

Substituting these back:

We have reached the summit. The question asks for $S_{30}-a_{40}$.

We have $S_{30}=13635$ and $a_{40}=2345$. Subtracting them gives:

We have navigated the pattern, derived the general term, performed the summation, and arrived at the answer. The final result is 11290.