Triangular Numbers 5 – Find N

We have that $\dfrac{n(n+1)}{2}=120$ . We can solve for $n$ : $\begin{aligned}\dfrac{n(n+1)}{2}&=120\\ n(n+1)&=240\\ n^2+n&=240\\ n^2+n-240&=0\\ (n-15)(n+16)&=0\\ n&= -16,15\end{aligned}$

Since we cannot have a negative triangular number, we discard the extraneous solution $n=-16$ . Thus, we are forced to conclude that $\boxed{n=15}$

We need to apply the formula $\frac{n(n+1)}{2}$ to find $n$ .

So, $\frac{n(n+1)}{2} = 120$

$n(n+1)= 120 \times 2$

$n^{2}+n=240$

$n^{2}+n-240=0$ (Solve with midde term splitting)

$n^{2}-15n+16n-240=0$

$n(n-15)+16(n-15) = 0$

$(n-15)(n+16)=0$

So, $n = 15,$ $-16$

But we cannot take negative value of $n$ because sum of negative numbers is not positive.

Check:

$\frac{15(15+1)}{2}=120$ $, n=15$

$\frac{15 \times 16}{2}=120$

$\frac{240}{2}=120$

$120=120$

Thus, the answer is: $\boxed{n = 15}$

­ $120 = \frac{n(n+1)}{2}$

$240 = n(n+1)$ We know the numbers are all natural numbers, so n and n+1 both have to be divisors of 240. $240 = 2^4 * 3 * 5$ If you look through all the possible combinations, the only ones that are one away from each other are $3*5 = 15$ and $2^4 = 16$ Therefore, $n = 15$ Keep in mind this won't work for algebra solutions where we aren't just dealing with counting numbers.

I dont know what $n ^ { \text { th } }$ means, but I will try my best.

A triangular number $T$ is equal to the sum of natural numbers, from $1$ to $a$ .

So, $120 = 1 + 2 + \cdots$ .

If we subtract the same numbers from each sides, then $119 = 2 + 3 + 4 + \cdots$ .

$117 = 3 + 4 + 5 + 6 + \cdots$ .

$114 = 4 + 5 + 6 + 7 + 8 + \cdots$ .

$110 = 5 + 6 + 7 + 8 + 9 + \cdots$ .

$105 = 6 + 7 + 8 + 9 + 10 + \cdots$ .

$99 = 7 + 8 + 9 + 10 + 11 + \cdots$ .

$92 = 8 + 9 + 10 + 11 + 12 + \cdots$ .

$84 = 9 + 10 + 11 + 12 + \cdots$ .

$75 = 10 + 11 + 12 + \cdots$ .

$65 = 11 + 12 + 13 + 14 + \cdots$ .

$54 = 12 + 13 + \cdots$ .

$42 = 13 + 14 + \cdots$ .

$29 = 14 + 15 + \cdots$ .

$15 = 15 + \cdots$ .

$0 = \cdots$ .

Hence, $\displaystyle 120 = \sum ^ { 15 } _ { n = 1 } n$ .

$120 = 1 + 2 + 3 + 4 + \cdots + 10 + 11 + 12 + 13 + 14 + 15$ .

So, 120 is the fifteenth triangular number.

$the \ first\ triangular\ number\ is\ : 1 \\ the\ second\ triangular\ number\ is\ : 1+2 \\ the\ third\ triangular\ number\ is\ : 1+2+3 \\ the\ nth\ triangular\ number\ is \\ : 1+2+3+4+...+n = \frac{n(n+1)}{2} \\ so\ n\ satisfy \\ \frac{n(n+1)}{2}=120 \\ n(n+1)=240=15 \times 16 \\ then\ n=15$

the first triangular number is : 1 the second triangular number is : 1+2 the third triangular number is : 1+2+3 the nth triangular number is : 1+2+3+4+...+n = n(n+1)/2 so n satisfy n(n+1)/2 =120 then n=15

$n(n+1)/2 =240\\n(n+1)=240\\n(n+1)=15*16\\n=15$ Hence answer is $15$ .

As we have seen that the no. of dots in nth fig is $\frac{n(n+1)}{2}$

So $\frac{n(n+1)}{2} = 120$

or $n=15$

I forgot the formula.

So, the thing I did is I added all the consecutive numbers till we get $120$ .

$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 120$

The last number is $15$ so the answer is $15$ .

Here is the code in Ruby:

for i in 1..20
    if (n * (n + 1)) / 2
        puts n
        break
    end
end