Question

Sum of natural numbers from 1 to n is 45 , then n =?

SOLVE BY FORMULA OF ARITHMETIC PROGRESSION​

Answer 1

The sum of natural numbers from 1 to 9 will be 45

Step-by-step explanation:

Given:

$s_{n}$= 45, a=1

To find:

value of n

Solution:

The sequence is in A.P.

Sum of the natural numbers from is 1 is 45, so a=1, $s_{n}$= 45,

The difference between the consecutive natural numbers will be 1

∴ d= 1

Now, $s_{n}$= $[(\frac{n}{2}) \X (2a+(n-1)d)]$

∴ $s_{n}$= $[\frac{n}{2}\ (2(1)+(n-1) 1)]$

$s_{n}$=$[\frac{n}{2}\ (2+(n-1) ]$

$s_{n}$= $[\frac{n}{2}\ (2+n-1)]$

$s_{n}$= $[\frac{n}{2}\ (1+n)]$

But, $s_{n}$= 45.........(given)

∴ 45= $[\frac{n}{2}\ (1+n)]$

∴ $\frac{45 (2)}{n}$= $(1+n)$

∴ $90= n(1+n)$

∴ $90= n+n^2$

∴ $n^2+n-90=0$

∴ $(n-9) (n+10)=0$...........(using factorization method)

∴ $n-9=0 \ or\ n+10=0$

∴ $n=9\ or\ n=-10$

But, n cannot be negative

Hence, n=9

∴ The sum of the first 9 natural numbers will be 45

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