if the sum of n successive odd natural numbers starting from 3 is 48. find the value of n.
Answers
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
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
⇰ + 2n = 48
⇰ + 2n - 48 = 0
Factorize it and we get,
⇰ + 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'.