If the property of the elementsĀ of the set B is 1+3+9+..+3^(n+1)=1093 what are the natural elements of the set.

1 Answer | Add Yours

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Let n be the elements of the set B.

To determine the elements of the set, we'll have to solve the sum of the consecutive terms of a geometric progresison, that represents the property of the elements of the set.

The number of terms of the geometric progression is n+2. The common ratio of the geometric progression is q = 3.

The sum of n+2 terms of the geometric progression is:

S = b1*(q^(n+2) - 1)/(q-1)

Foe b1 = 1 and q = 3, we'll get:

S = [3^(n+2) - 1]/(3-1)

S = [3^(n+2) - 1]/2

But S = 1093 => [3^(n+2) - 1]/2 = 1093

Therefore, we'll have:

3^(n+2) - 1 = 2186

3^(n+2) = 2187

We'll create matching bases since 2187 = 3^7:

3^(n+2) = 3^7

Since the bases are matching, we'll apply one to one property:

n + 2 = 7

n = 5

The natural number that has the given property of the set B is n = 5.

We’ve answered 318,989 questions. We can answer yours, too.

Ask a question