Question

2481 = (±1±2±3± ………±????) find the minimum value of n.

Answer 1

Answer:

Step-by-step explanation:

sum of n terms = n(n+1)/2

⇒2481=n(n+1)/2

⇒4962=n²+n

⇒n²+n-4962=0

by solving this

n=6.5 which is not possible

Answer 2

Answer:

The value of n is 70

Step-by-step explanation:

Step 1:

Let’s take only the positive values

1 + 2 + 3 + .............................+ n

Step 2:

Sum of first n Terms = n(n + 1) / 2  

n(n + 1)/2 ≥ 2481

=> n(n+1) ≥ 4962

70 x 71 = 4970 > 4962

Step 3:

Sum of 70 numbers = 2485   (4 above than 2481)

So, 4 is extra, to reduce by 4 (we can take 2 as - instead of +)

If we take +2 as - 2 and all other terms positive

Step 4:

Then the sum would be exactly as required = 2481

Therefore n = 70

If we put Question like this

1 - 2 + 3 - 4 + ...........................................+ n

1 -2 = -1 3 -4 = -1 and so on

Step 5:

Then for 2k terms sum would be – k

Step 6:

Then last term should be 2k+1 such that sum 2481

=> -k + 2k + 1 = 2481

=> k = 2480

=> 2k = 4960

2k+1 = 4961

Last term = 4961