Question

Find such numbers from 1 to 288 natural numbers. So that the sum of a small number is equal to the sum of a large number.​

Answer 1

Answer:

Step-by-step explanation:Let the number be n.

So:

1+2+3................n-1 = n+1+n+2......................288

In the first AP,

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

no. of terms=n-1

a=1

d=1

Sn=(n-1)/2*[2+n-1-1]

=(n-1)/2*[n]................(1)

Second AP

a=n+1

d=1

no. of terms=288-n

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

Sn=(288-n)/2*[2+288-n-1]

=(288-n)/2*[n+289]..............(2)

(1)=(2)

So:

(n-1)/2*[n]=(288-n)/2*[n+289]

Multiplying both sides by 2.

==) n(n-1)=(288-n)(289+n)

==)n²-n=83232+288n-289n-n²

==)2n²-n=83232-n

==)2n²=83232

==)n²=83232/2

==)n²=41616

==)n=204

n cannot be negative so n is not -204

The term hence is 204

Hope it helps you.

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Answer 2

We can put what Gauss discovered into an easy-to-use formula, which is: (n / 2)(first number + last number) = sum, where n is the number of integers.

We seek n∈{1,…,288} for which

1+2+3+⋯+(n−1)=(n+1)+(n+2)+(n+3)+⋯+288,

or for which

2(1+2+3+⋯+(n−1))+n=1+2+3+⋯+288.

Thus,

n2=n(n−1)+n=12⋅288⋅289=122⋅172 ,

implying n=12⋅17=204. ■