Question

from natural numbers 1 to 288 , find the number for which the sum of numbers smallest than that numbers equal the sum of numbers greater than that number.

Answer 1

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.

Answer 2

Hlo mate :-

Solution :-

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● We seek n∈{1,…,288}n∈{1,…,288} for which

1+2+3+⋯+(n−1)=(n+1)+(n+2)+(n+3)+⋯+2881+2+3+⋯+(n−1)=(n+1)+(n+2)+(n+3)+⋯+288,

☆or for which

2(1+2+3+⋯+(n−1))+n=1+2+3+⋯+2882(1+2+3+⋯+(n−1))+n=1+2+3+⋯+288.

Thus,

n2=n(n−1)+n=12⋅288⋅289=122⋅172n2=n(n−1)+n=12⋅288⋅289=122⋅172,

implying n=12⋅17=204

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