If the nth term of a sequence Is given by 2*3^n-2, find the first three term of the sequence
To find the first three terms of the sequence, we plug in n=1, n=2, and n=3 into the given nth term formula and simplify:
- When n=1: 2*3^(1-2) = 2*3^-1 = 2/3
- When n=2: 2*3^(2-2) = 2*3^0 = 2
- When n=3: 2*3^(3-2) = 2*3^1 = 6
Therefore, the first three terms of the sequence are 2/3, 2, and 6.
To find the first three terms of the sequence, we substitute the values of "n" into the formula for the nth term of the sequence, which is given as 2 * 3^(n-2).
Let's calculate them step by step:
For n = 1:
2 * 3^(1-2) = 2 * 3^(-1) = 2 * (1/3) = 2/3
So, the first term of the sequence is 2/3.
For n = 2:
2 * 3^(2-2) = 2 * 3^0 = 2 * 1 = 2
So, the second term of the sequence is 2.
For n = 3:
2 * 3^(3-2) = 2 * 3^1 = 2 * 3 = 6
So, the third term of the sequence is 6.
Therefore, the first three terms of the sequence are 2/3, 2, and 6.