What is the number of the terms of geometrical sequence if its first, fourth and the last term is equal to 2, 54 and 486 respectively.

Answer 1

This problem should first be solved assuming an integer solution. If none is found, then we should pursue an algebraic approach. Both will be presented.

Assuming an integer solution, not that 54 is not a power of 2. Hence, the common factor between neighboring terms cannot be 2 or any power of 2. Let's try a common factor of 3:

1st term = 2

2nd term = 2*3 = 6

3rd term = 6*3 = 18

4th term = 18*3 = 54

Hence the common ratio is 3.

let's continue:

5th term = 54*3 = 162

6th term = 162*3 = 486

Hence, the answer is 6

Note: using -3 as a common ratio should be checked, but then the 4th term would be -54.

An algebraic solution is:

Assume r to be the common ratio.

1st term = 2

2nd term = 2r

3rd term = 2r²

4th term = 2r³ = 54

Solving for r leads to

r³ = 27

r = 3

The nth term of this geometric progression is = 2*r^(n-1) = 2*3^(n-1)

We must solve for n in

2*3^(n-1) = 486

3^(n-1) = 243

Factoring 243 leads to 243 = 3*81 = 3*9*9 = 3*3²*3² = 3⁵

Hence

n - 1 = 5

n = 6.