Birthday Leaves

$\text{ the rule of the sum of natural numbers will work here.}$

$\color{#BA33D6}n o w,$ $\color{#BA33D6} \frac{n(n+1)}{2}=276$

$\color{#302B94}o r,$ $\color{#302B94} n^2+n=552$

$\color{#20A900} o r,$ $\color{#20A900}n^2+n-552=0$

$\color{#EC7300}o r,$ $\color{#EC7300}(n-23)(n+24)=0$ ....................[middle term]

$\color{#D61F06}or,$ $\color{#D61F06}n=23$ .............................[n can not be negative]

Let us consider this a series starting with 1, going to n terms. Where the sum of n terms is 276

This is the sum of natural numbers. According to the rules of progressions, the sum of natural numbers are $\frac{n(n+1)} {2}$

Therefore $\frac{n(n+1)} {2} = 276$ or $n^{2} + n - 552 = 0$ Factoring, we get $(n-23) (n+24) = 0$ or $n = 23 / n = -24$ But since 'n' can't be a negative number,

$\boxed{n=23}$

On the 23rd.

276 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23.

Just use the listing method :P

AS the series is 1, 2, 3, ... n

and 1 + 2 + 3 + ... + n = 276

sum = n(n + 1)/2 = 276 or n² + n = 552 or n² + n – 522 = 0

(n – 23)(n + 24) = 0 taking + ve value n = 23

Let up solve this using AP. Here,1,2,3,4,5......... a=1(first term) d=1(common difference) Sn=276(sum of all terms) n=?

Sn=n/2 [2a+(n-1)d] =n/2 [2+n-1] =n/2*[n+1] = n^+n/2=276 = n^2 + n = 552 =n^2+ n - 552=0 Therefore, n^2+n-552=0

On solving the above quadratic equation we get n= -24 or 23 As n can't be negative Hence.n=23 So birthday falls on 23rd of the month.