Perfect AP

$\textbf{Solution.}$ We will prove it in a more general case,let us take the $1$ st term to be $k^{2}$

When $n=1$ ,the sum of the first term is a perfect square = $k^{2}$ .

The sum upto the $n$ -th term is $S_n = \frac{n}{2}[2a+(n-1)d]$ ,since these are all perfect squares we have $[2a+(n-1)d] > 0$ and hence $d>0$

Now let us choose any odd prime $p$ . $S_p = \frac{p}{2}[2a+(p-1)d]$

Since $p|S_p$ and $S_p$ is a perfect square, $p^{2} | S_p$ , $p|(2a-d)+pd$ $p|(2a - d)$ This relation is true for all primes $p$ ,but since $(2a-d)$ is finite,we must have $2a - d = 0$ and hence $d = 2a = 2k^{2}$

So the series is ${k^{2},3k^{2},5k^{2},.......}$ .

In other words,the $n$ -th term is $t_n = (2n-1)*k^{2}$ .

In this problem clearly $k = 6$ ,So the $100$ -th term of the series will be = $(2*100-1)*6^{2} = \boxed{7164}$

A relatively well-known fact about the AP of a sequence of odd numbers is that the sum is always a perfect sqaure: 1, 1+3 = 4, 1+3+5=9 etc then we will just have to match the first term of this to the one of the question by multiplying each term by 36

Since 36 is a perfect square this property is preserved: $\text{for } a_{n}=2n-1 \quad \text{where } n\in\mathbb{Z}\colon \quad \sum_{k=1}^{n}a_{k}=n^2$ then $36\sum_{k=1}^{n}a_{k}=36n^2=6^{2}n^{2}=(6n)^{2}$ which is still a perfect square

So now it reduces to finding the 100th term of this new sequence: $a'_{100}=36a_{100}=36(2\times 100 - 1) = \boxed{7164}$ The proof of sum of odd numbers equal to a perfect square can be trivially proven by induction or graphically and hence is omitted here.

I did it this way:

S(n) = $\frac{n}{2}$ [2a+(n-1)d] = $m^{2}$ ,where m is some positive integer.

Rearranging terms in L.H.S,

( $\frac{d}{2}$ ) $n^{2}$ + (a- $\frac{d}{2}$ ) n = $m^{2}$

The only way to complete the square on the L.H.S is by making d=2a=72,

and so the sum of first n terms will be of the form S(n)= $m^{2}$ = $(6n)^{2}$ .

Hence the 100th term will be a+(100-1)d= $\boxed{7164}$ .

This is my solution (Sorry, I don't know how to use LaTeX here): The terms of this AP are 36, 36+a, 36+2·a, ... So the sum of the first n terms is: S n = 36 + (36+a) + (36+2·a) + ... + (36 + (n-1)·a) = (36 + ... + 36) + (0·a + 1·a + 2·a + ... + (n-1)·a) Therefore, S n = 36·n + a·n·(n-1)/2 If we expand that expression, we have S n = 36·n - (a/2)·n + (a/2)·n^2 = (36 - a/2)·n + (a/2)·n^2 So one realizes that it would be enough if a/2 were 36, since S n would be a perfect square: S_n = 36·n^2. Hence, we are looking for a = 72. With that, the 100th term of the AP is 36 + 99·72 = 7164

Let d be the common difference of the AP. We want that $\sum_{i=1}^{k} a_k = square ,$ for any $k \in \lbrace 1,2,...,100 \rbrace$ .Is simple to see that the sum is $36k + d \frac{k(k+1)}{2}$ , and so we want

$36k + d \frac{k(k+1)}{2} = square$ for any $k$ .

Rearranged the sum like $36( k + d \frac{k(k+1)}{72})$

and see that if $d= 72$ we have

$\sum_{i=1}^{k} a_k = 36k^2$ for any k.

So $d=72$ and $a_{100}= 36+99 \times 72 = \boxed{1764}$

First of all, nothing in the problem description says the constant sequence $a_n = 36$ $ cannot be a solution. Secondly, the solution is best done by drawing. Draw a 1x1 square. To get a 2x2 square, you add 3. To get a 3x3 square you add 5. Conclusion: the sum of the first $n$ odd numbers is a perfect square. So let $a_n=2n-1$ . However, we want our sequence to start with 36, with is a perfect square, so we multiply $a_n$ with 36 to get $b_ n = 72n-36$ . This is our sequence. It works, because the product of a square and a square is a square.