1000th number in a triangle

Realise that the total number of numbers read out after N rows is the sum of the series

$1+2+3....+n$ , i.e the sum of the triangle numbers, which is an arithmetic series, which has the summation formula:

$S_{N}=\frac{N(N+1)}{2}$

So we can work out what row the 1000th number is in by finding the N such that $S_{N}=1000$

Meaning:

$N^{2}+N-2000=0, N=\frac{-1±\sqrt{8001}}{2}$

$90^{2} =8100 => \sqrt{8001} \sim 89 => N \sim 44$

So 1000 is part of the way into the 45th row of the triangle. There are 990 numbers up to the 44th row, implying that the 1000th number read is 10 into the 45th row, so is 10.

We first find the row. The sum of the number of values of the first $n$ rows is the sum of the integers representing those rows; thus the total number of values in rows $n$ and above is $\binom{n+1}{2}$ . We are then looking for the closest row to the $1000^{th}$ number, so we approximate this by $\binom{n+1}{2}\approx1000\rightarrow n^2-n\approx 2000 \rightarrow n^2-n-2000\approx0$

We then use the quadratic formula and take the positive solution to get $n \approx \frac{1+3\sqrt{889}}{2}\approx 44$ (note that even though we have taken $\lfloor \sqrt{889} \rfloor$ along with our original approximation of $\binom{n+1}{2}\approx 1000$ , the second approximation does not affect the first and vice versa, meaning our approximation is still accurate.) This implies the last complete row before the $1000^{th}$ number is $44+1=45$ . We have $\binom{45}{2}=990$ , so the last value of row $45$ is also the $990^{th}$ number.

Thus the $1000^{th}$ number is the $10^{th}$ number of row $46$ , or $\boxed{10}$

To find what row the 1000th number is in, you want to find the term in the sequence $1 + 2 + 3 + 4 \ldots$ that puts you over $1000$ , which can be done using the sequence's explicit formula, $\frac {X(X+1)}{2}$ . If you solve the formula for $1000$ , you'll find that $X=44.224\ldots$ , showing that the 1000th number will be in the 45th row. If we put $X=44$ into the formula, we get $990$ , showing that the 1000th number is 10 terms into row 45, and the 10th term in any row will be $\boxed{10}$

in row 1, we have 1 integer (1)

in row 2, we have 2 integers (1,2)

in row 3, we have 3 integers (1,2,3)

......

in row 1000, we have 1000 integers (1,2,3,...,1000)

So, take the number of integers in row n to form an arithmetic progression (AP)

AP: 1,2,3,...,1000

To determine the row of 1000th number,

S <= 1000

(n/2) [2a+(n-1)d] <=1000

(n/2) [2x1+(n-1)1] <=1000

(n/2) (2+n-1) <=1000

(n/2) (n+1) <=1000

n^2+n<=2000

n^2+n-2000<=0

n=44.23 or -45.22

n denotes the number of row, so n cannot b negative, therefore n=44.23

This n value represents 1000th number is in the 45th row

To determine where the 1000th number in the 45th row, substitute n=44 into

(n/2) (n+1)=990

1000-990=10

The 1000th number is the 10th element in the 45th row, therefore the 1000th number is 10

this is a "Pattern Recognition" group sequence, let transform the problem to: what is the value at 1000th term, of the sequence: (1), (1,2), (1,2,3), ...

First: we find the row contains the 1000th term by using n(n+1)/2 = 1000 n = 44, so the 1000th term in group 44 + 1 = 45

Second: max value of group 44 is 44*45/2 = 990 then in group 45, the value of 1000th is 10 (because 1000 - 990 = 10)

The rightmost integer on each row will increase by one each time, but their positions are the triangular numbers. Since 990 is a triangular number, it will be on the end of a row. 1000 is 10 more than 990, so the 1000th number will be 10.

The first $n$ rows of the triangle contain $r_n = \frac{n^2 + n}{2}$ numbers. Setting $r_n = 1,000$ and solving for $n$ , we see that $r_{44} < 1,000 < r_{45}$ and so the $1,000^{\text{th}}$ number must be in the $45^{\text{th}}$ row. Now we just need to solve $r_{44} + x = 1,000$ which gives us $x = 10$ .

the triangle will be formed as 1 1 2 1 2 3 and so on. since no. of numbers in each row are increasing by 1... the 10th no. of 45th row will be the 1000th no. and by the way the triangle is formed it will be 10

numbers of integer at row $i$ = $i$

At row 1, we got 1 integer(s); at row 2, we got 2; at row 3 we got 3....

This form an A.P. with $a=1$ and $d=1$

For the 1000th integer, we must find the number of row containing it.

From sum of A.P formula, $\frac{n}{2} (1 + n) = 1000$ ,

solving it and we get $n = 44.22$ or $n = -45.22$ ,

reject the negative answer since in n is always positive.

Hence, we get $n = 44.22$ , meaning that the 1000th integer is at row 45.

Up to row 44, the number of integers is $1 + 2 + 3 + ... + 44 = 990$ and it's 10 more to go to the 1000th integer.

Hence, the 1000th integer(or 10th in row 45) is $10$

The total of the numbers in a triangle with $n$ rows is $\frac{(n+1) \times n}{2}$ .So the highest value of $n$ where $\frac{(n+1) \times n}{2} < 1000$ is 44 where $\frac{(n+1) \times n}{2}$ is 990.So the 1000th number is the 10th number in row 45 which is 10.

Let the 1000 th number that is read be in line no. N . then N( N+1)/2 < 1000 The max such N is 44 and N ( N+1)/2 is 990 thereby the 1000 th number is 10

Use Gauss law to find that sum of integers from 1 to 45 is 990. Ans 10