Question

If sum of first n even natural number is 420 then find n

Answer 1

Hi !

Even natural no:s form an A.P :-

2,4,6,8 .....

a = 2 , d = 2 , Sn = 420

Sn = n/2 [ 2a + (n-1)d]
420 = n/2 [ 4 + 2n - 2 ]
840 = n [ 2 + 2n ]

 2n²+ 2n = 840

2 [ n² + n ] = 840

n² + n = 420

n² + n - 420 = 0
               -------------
n = -b ± √ b² - 4ac  /   2a
     
   = -1 ± √1 - 4 x 1 x -420 = 2
 
    = -1 ± √1681 / 2
 
   = -1 ± 41 /2
 
   = -1 + 41/2 = 40/2 = 20

n = 20

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Here, we have

2 + 4 + 6 + 8 + ..... to n terms = 420

We know that,

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

According to the Question,

⇒ n/2[2 × 2 + (n - 1) × 2] = 420

⇒ n(2 + n - 1) = 420

⇒ n(n + 1) = 420

⇒ n² + n = 420

⇒ n² + n - 420 = 0

By using factorization method, we get

⇒ n² + n - 420 = 0

⇒ n² + 21n - 20n - 420 = 0

⇒ n(n + 21) - 20(n + 21) = 0

⇒ (n + 21) (n - 20) = 0

⇒ n + 21 = 0 or n - 20 = 0

⇒ n = - 21, 20 (As n can't be negative)

⇒ n = 20

Hence, the value on n is 20.