Question

Sum of first ‘n' natural numbers is 231. What number is ‘n'?

Answer 1

Given:

• Sum of n natural numbers is 231.

To Find:

• The value of n .

Answer:

Given that the sum of first n nautural numbers is 231.

So , we know a formula for sum of first n nautural numbers as ,

$\large{\underline{\boxed{\bf{\leadsto S_{n}=\dfrac{n(n+1)}{2}}}}}$

⇒ n(n+1)/2 = 231.

⇒n(n+1) = 231 ×2 .

⇒ n² + n = 462 .

⇒ n² + n - 462 = 0 .

⇒n² + 22n - 21n - 462 = 0.

⇒n( n + 22 ) - 21( n +22 ) = 0.

⇒ ( n + 22 ) ( n - 21 ) = 0 .

⇒ n = 21 , -22 .

Since n can't be negative ;

Hence n = 21 .

Hence the value of n is 21.

Answer 2

Given ,

• Sum of first ‘n' natural numbers is 231

We know that , the sum of first n natural number is given by

S = n(n + 1)/2

Thus ,

$\sf \mapsto 231 = \frac{n(n - 1)}{2} \\ \\ \sf \mapsto 462 = {(n)}^{2} - n \\ \\ \sf \mapsto {(n)}^{2} - n - 462 = 0 \\ \\ \sf \mapsto \sf \mapsto {(n)}^{2} - 22n + 21n - 462 = 0 \\ \\ \sf \mapsto n(n - 22) + 21(n - 22) = 0 \\ \\ \sf \mapsto (n + 21)(n - 22) = 0 \\ \\ \sf \mapsto n = - 21 \: \: or \: \: n = 22$

Since , number of terms can't be negative

$\sf \therefore\underline{The \: number \: of \: terms \: (n) \: whose \: sum \: is \: 231 \: is \: 22}$