Question

the sum of first n-natural numbers is 820 then find the value 0f 'n '?​

Answer 1

Answer:

n(n+1)/2 = 820

n(n+1) = 820×2

=1640

(n^2)+n = 1640

the nearest square to 1640 is 1600

I.e., n^2 = 1600

and n= 40

Answer 2

Answer:

The sum of first n-natural numbers is 820.

Step-by-step explanation:

Natural numbers are the set of positive integers starting from one to infinity.

Example: 1,2,3,4,5,6,,,,,,,,,,,,,,,∞

Here in this problem the sum of n- natural numbers is 820.

This is a problem of quadratic equation in two variable.

So,

let the numbers be n, (n+1),(n+2),,,,,,,,,and so on.

sum of n - natural numbers is:

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

now,

$S_{n} = \frac{n(n+1)}{2}=820$

$n(n+1)=1640$

$n^{2} +n-1640=0$

$n^{2} +41n-40n-1640=0$

$n(n+41) -40(n+41)=0$

$(n-40)(n+41)=0$$n=40, n=-41$

Since natural numbers cannot be negative so -41 is not the solution of the quadratic equation and hence 40 is the correct answer.

The sum of first 40 natural numbers will give the result as 820.

the correct value of n is 40.

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