What is the 11th term of the sequence 2, 8, 32, 128 A. 262,144

B. 524,288

C. 1,048,576

D. 2,097,152

Step-by-step explanation:

Given the sequence

2, 8, 32, 128..

The sequence is a geometric sequence. The nth term of the geometric sequence is expressed as;

Tn = arⁿ⁻¹

a is the first term

r is the common ratio

n is the number of terms

From the sequence;

a = 2

r = 8/2 = 32/8 = 4

n = 11 (since we are to find the 11th sequence)

Substitute into the formula;

T₁₁ = 2(4)¹¹⁻¹

T₁₁ = 2(4)¹⁰

T₁₁ = 2(1,048,576)

T₁₁ = 2,097,152

Hence the 11th term of the sequence is 2,097,152

c.

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.