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Write the first six terms of the geometric sequence with the first term, \(a_{1}\), and common ratio, \(r\). \(a_{1}=-2, r=-3\)

Short Answer

Expert verified
The first six terms of the geometric sequence with \(a_{1} = -2\) and \(r = -3\) are -2, 6, -18, 54, -162, 486

Step by step solution

01

Understanding the formula

The formula to find the n-th term of a geometric sequence is \(a_{n} = a_{1} \cdot r^{(n-1)}\), where \(a_{1}\) is the first term, \(r\) is the common ratio, and \(n\) is the term number. In this situation, \(a_{1} = -2\) and \(r = -3\).
02

finding the first term

The first term is simply the given starting term \(a_{1}\), which is -2 in this case.
03

finding the second term

By substituting \(n = 2\), \(a_{1} = -2\), and \(r = -3\) into the formula, the second term (\(a_{2}\)) can be calculated as follows: \(a_{2} = a_{1} \cdot r^{(n-1)} = -2 \cdot (-3)^{(2-1)} = -2 \cdot -3 = 6\).
04

finding the third term

Substitute \(n = 3\), \(a_{1} = -2\), and \(r = -3\) into the formula, the third term (\(a_{3}\)) can be calculated as follows: \(a_{3} = a_{1} \cdot r^{(n-1)} = -2 \cdot (-3)^{(3-1)} = -2 \cdot 9 = -18\).
05

finding the remaining terms

Similarly, by substituting into the formula for the \(n = 4, 5, 6\) terms, it's found that: \(a_{4} = a_{1} \cdot r^{(n-1)} = -2 \cdot (-3)^{(4-1)} = 54\); \(a_{5} = a_{1} \cdot r^{(n-1)} = -2 \cdot (-3)^{(5-1)} = -162\); \(a_{6} = a_{1} \cdot r^{(n-1)} = -2 \cdot (-3)^{(6-1)} = 486\).

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Most popular questions from this chapter

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