Fun with sums

Let the first term be $a$ and the common difference be $d$ .
Let the sum of the first five terms be $S_5 = \dfrac{5}{2} \left(2a + \left(5-1\right)d\right)\\ 40 = \dfrac{5}{2} \left(2×2 + 4d\right)\\ \dfrac{40 × 2}{5} - 4 = 4d\\ 12 = 4d\\ d = \dfrac{12}{4}\\ = 3$ .

Let there be $n$ terms in the progression, then $101$ is the $n^{th}$ term. So, $101 = a + (n - 1)d\\ 101 = 2 + (n-1)×3\\ \dfrac{99}{3} = n-1\\ n = 34$ .
So, there are $34$ terms in the given progression.

Since the last $5$ terms whose sum we have to find is also a part of the arithmetic progression, they would have the same common difference. So, the sum of the last $5$ terms = $\dfrac{5}{2} \left(a_{30} + a_{34}\right)$ where $a_{30}$ is the fifth last term that is the $\left(34 - 5 + 1= 30\right)$ th term and $a_{34}$ is the last term.

So, Sum of last 5 terms = $\dfrac{5}{2} \left(a + 29d + 101\right)\\ = \dfrac{5}{2} \left(2 + 29×3 + 101\right)\\ = \dfrac{5}{2} × 190\\ = \boxed{475}$ .

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Alternate method:-
We know that last term is $101$ and common difference is $3$ . So, the sum of last 5 terms would be $101 + 98 + 95 + 92 + 89 = \boxed{475}$ .

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Alternate method:-
Find the sum of all the terms and subtract from it the sum of first (n-5) terms.
Sum of all terms = $\dfrac{34}{2} (2 + 101) = 1751$
Sum of first (n-5 = 29) terms = $\dfrac{29}{2}(2a + 28d) = \dfrac{29}{2}(2×2 + 28×3) = 1276$
So, sum of last 5 terms = $1751 - 1276 = \boxed{475}$ .

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Alternate method:-
This is the same as the second method but observation based.
Common difference is 3 and first term is 2. So, observe that nth term is of the form 3n-1. So, the sum of last 5 terms would be $(3×34-1)+(3×33-1)+(3×32-1)+(3×31-1)+(3×30-1)\\ = \left(3×\displaystyle \sum_{n=30}^{34} n\right) - 5\\ = (3×160) - 5\\ = \boxed{475}$ .

The average number for the first five terms is $\frac{40}{5} = 8$ which is 6 more than $a_1 = 2.$

Similarly the average term for the last five term must be $101 - 6 = 95$ , and their sum must be $95\cdot 5=475$

For an AP, $a_r+a_{n-r+1}$ sum to the same value, i.e. $103$ in this case.

Hence sum of the last five terms is $5(103)-40 = 475$ .