Question

if the sum of n successive odd natural numbers starting from 3 is 48. find the value of n.

Answer 1

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Here,

3 + 5 + 7 + 9 + .... n = 48

Using Formula we have,

S(n) = n/2[2a + (n - 1)d]

Situation,

⇒ n/2[2 × 3 + (n - 1) × 2] = 48

⇒ n(3 + n - 1) = 48

⇒ n² + 2n - 48 = 0

Factorize it,

⇒ n² + 8n - 6n - 48 = 0

⇒ n(n + 8) - 6(n + 8) = 0

⇒ (n + 8)(n - 6) = 0

⇒ n + 8 = 0 or n - 6 = 0

⇒ n = - 8, 6 (Negative value is not applicable)

Hence

⇒ n = 6

Therefore,

Required value on n = 6

Answer 2

Answer

According to the question,

3 + 5 + 7 ........ = 48

Here it is given sum of successive odd number starting from 3 is 48.

We know that two successive odd natural number differ by '2' .For example : 7 - 5 = 2 , 9 - 7 = 2

Clearly this is in A.P

So by formula,

Sn = n/2 (2a + (n-1)d )

⇰ n/2 ( 2 × 3 + ( n -1 ) 2 ) = 48

⇰ n/2 (6 + 2n -2 ) = 48

⇰ n/2 (2n + 4 ) = 48

⇰ $\rm n^2$ + 2n = 48

⇰ $\rm n^2$ + 2n - 48 = 0

Factorize it and we get,

⇰ $\rm n^2$ + 8n - 6n - 48 = 0

⇰ n(n + 8) - 6(n + 8) = 0

⇰ (n+8)(n - 6) = 0

Now we got,

• n = -8
• n = 6

Here in this condition we can have negative number as 'n'.

So n = 6