Question

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

Answer 1

Here

a=3

d=2

n=?

Sn=48

We know that Sn=n/2[2a+(n-1)d]

48=n/2[2×3+(n-1)2]

48×2=n[6+(n-1)2]

96=n(6+2n-2)

96=n(4+2n)

96=4n+2n^2

4n+2n^2-96

2n^2+4n-96

n^2+2n-48

n^2+8n-6n-48

n(n+8) - 6(n+8)

(n+8)(n-6)

n=-8, +6

n cannot be negative ,therefore n = 6

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Here, we have

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

We know that

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

Using that, where a = 3 and d = 2

According to the Question,

⇒ n/2[2 × 3 + (n - 1) × 2] = 48

⇒ n(3 + n - 1) = 48

⇒ n² + 2n - 48 = 0

By using factorization method, we get

⇒ 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 (As n can't be negative)

⇒ n = 6

Hence, the required value on n is 6.