Question

The sum of natural numbers from 1 to n is 36. find the value of n.​

Answer 1

Answer:

n=8

Step-by-step explanation:

Sum of natural number is given by

$S_{n}=\frac{n}{2} [2a+(n-1)d]$

here a=1, d=1

$S_{n}=\frac{n}{2} [2a+(n-1)d]\\36=\frac{n}{2} [2\times1+(n-1)1]\\72=2n+n^2-n\\n^2+n-72=0\\(n+9)(n-8)=0\\n=-9 , n=8$

So n is not equal to 9 because n would not be negative.

Therefore,n=8

Answer 2

Question :-

The sum of natural numbers from 1 to n is 36 ,then find the value of n .

Answer :-

→ Here n = 8 ,

Solution :-

According to the question,

→ 1 , 2 , 3 ......n

this is a series arithmetic progression (AP)

Given that sum of these terms is 36 ,

Here ,

• First term (a) = 1

• Last term (l) = n

$\: \: \: \: . \sf \: S_n = 36$

We know that ,

Sum of n terms -

$\to \sf S_n = \frac{n}{2} (first \: term \: + last \: term) \\ \\ \to \sf \: 36 = \frac{n}{2} (1 + n) \\ \\ \to \sf \: 72 = n + {n}^{2} \\ \\ \to \sf \: {n}^{2} + n - 72 = 0 \\ \\ \sf \: split \: the \: middle \: term \: \\ \\ \to \sf \: {n}^{2} + 9n - 8n - 72 = 0 \\ \\ \to \sf \: n(n + 9) - 8(n + 9) = 0 \\ \\ \to \sf (n + 9)(n - 8) = 0 \\ \\ (1) \sf \: if \: n + 9 = 0 \\ \\ \longmapsto \boxed{\sf \: n = - 9} \\ \\ (2) \sf \: if \: n - 8 = 0 \\ \\ \longmapsto \boxed{\sf n = 8} \\ \\ \because \bf sum \: of \: natural \: numbers \: would \: not \:be \: \\ \bf \: negative \: \\ \\ \sf \: hence \: \: n = 8 \\ \\ \sf \: sum \: of \: 8 \: natural \: numbers \: = 36 \\ \\ \sf \: these \: numbers \: are \: \\ \to \: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$