The problem statement
1, -3, 4, 1, -3, 4,1, -3, 4,.....
In the sequence above, the first 3 terms repeat without end. What is the sum of the terms of the sequence from the 150th term to the 154th term?
I know the naive approach to solve this problem.
Can I express this sequence in terms of geometric progression or arithmetic progression?
On
All you need is which of the three numbers $1,-3,4$ is the $150^{th}$ term. A look at the start shows that the 3rd,6th,9th etc. are all 4. That means, if $k$ is a multiple of $3,$ then the $k^{th}$ term of the sequence is a $4.$ Now you want five (not four as I had before) terms of the sequence starting at the 150-th term. And since $150=3 \cdot 50$ we know the 150-th term is a $4.$ So given the pattern which repeats, the four terms are: $4,1,-3,4,1.$
Here's how to solve adding the $k$th term to the $m$th term in general.
To figure out any $N$th term write $N$ as $3q + r$ where $r$ is the remainder when you divide $N$ by $3$. If $r = 1,2$ or $0$ then $N$th term is $1, -3$ or $4$ respectively. This will work to figure out the $k$th term, the $m$th term or any term in-between.
The $k$th to $m$th term will by $(m+1) - k$ terms in total. $1 + (-3) + 4 = 2$ so every three terms add to $2$. So the group of $k$ to $m$th terms will have a full set of $\lfloor \frac {(m+1) - k}3 \rfloor$ of these groups of $3$ terms. These will add to $2\times \lfloor \frac {(m+1) - k}3 \rfloor$.
But there might be one are two more terms if $3$ does not divide evenly into $(m+1) - k$.
So.....
Every integer $N$ can be written as $3q + r$ where $r$ is the remainder and $q =\lfloor \frac N3\rfloor$.
Let $(m+ 1)-k = 3q + r$.
If $r = 0$ the sum will be $2\times q$.
If $r =1$ the sum will be $2\times q$ plus one extra term. Consider that the last term. Call that $a_m$ and we determine the $m$th term as we did above: Write $m= 3Q + R$ and if $R = 1,2,0$ then $a_m= 1,-3,4$.
So if $r = 1$ the sum will be $2\times q + a_m$.
And if $r =2$ we will have two extra terms and the sum will be $2\times q + a_m + a_{m-1}$.
.....
This is WAY overkill for your problem but we could solve $a_{150} + .... +a_{154}$ by noting that there are $(154+1)-150 = 5$ terms. That is one full triplets of terms that add to $2$.
The two extra terms are $a_{154}$ and $a_{153}$. $154 = 3*51 + 1$ so $a_{154} = 1$ and $a_{153}$ is the previous term $4$. So the sum is $2 + 1 + 4 = 7$.
....
To see this clearly. We can figure the $150$th term is $\underbrace{1,-3,4}_3,\underbrace{1,-3,4}_6,.......,\underbrace{1,-3,4}_{147}, \underbrace{1,-3,4}_{150}$ so the $150$th term is $4$.
And so the sum of the $150$th to $154$th term is
$(4 + 1 + -3) + 4 + 1 =$
$[2] + 4 + 1 = 7$.
......
but if say, we were asked to find the sum of the $213$th term to the $1024$ term we could figure:
There are $(1024 + 1) -213 = 812$ terms. $812 = 3\times 270 + 2$. So there are $270$ triplets adding to $270\times 2 = 540$. There are two more terms. As $1024 = 3\times 341 + 1$ then $a_{1024} = 1$ and $a_{1023} = 4$ so the total is $540 + 1 +4 = 545$.
To verify and $213$ is divisible by $3$ we now then $213$ term is $4$ and our sum is:
$\underbrace{(4 + 1 -3) + (4+1 -3) + ........ + (4+1-3)}_{(213,214,215),(216,217,218)......,(1020,1021,1022)} + 4 + 1=$
$\underbrace{2 + 2+..... + 2}_{3\times 71,..... ,3\times 340} + 5=$
$\underbrace{2 + 2 + .... + 2}_{(340+1)-71\text{ times}}+ 5 =$
$2\times 270 + 5 = 545$.