Question

the first four terms of an arithmetic sequence are given by ( a_1 = 6 ), ( a_3 = 18 ), and ( a_4 = 24 ). which formula can be used to find ( a_n )?
a. ( a_n = 6 cdot 2^{n - 1} )
b. ( a_n = 6 cdot 2^n )
c. ( a_n = 6n )
d. ( a_n = 6 + 6n )
e. ( a_n = 12 - 6n )

Explanation:

Step1: Recall Arithmetic Sequence Formula

The formula for the \(n\)-th term of an arithmetic sequence is \(a_n=a_1+(n - 1)d\), where \(a_1\) is the first term and \(d\) is the common difference. Given \(a_1 = 6\), we need to find \(d\). From \(a_1 = 6\), \(a_3=18\), using the formula \(a_3=a_1 + 2d\), substitute values: \(18=6 + 2d\). Solve for \(d\): \(2d=18 - 6=12\), so \(d = 6\).

Step2: Derive the Formula for \(a_n\)

Substitute \(a_1 = 6\) and \(d = 6\) into the arithmetic sequence formula: \(a_n=6+(n - 1)\times6\). Simplify: \(a_n=6+6n-6=6n\). We can also check with other terms. For \(n = 4\), \(a_4=6\times4 = 24\), which matches the given value. Now check other options:

• Option A: \(a_n=6\cdot2^{n - 1}\) is a geometric sequence formula, not arithmetic. For \(n = 1\), \(6\cdot2^{0}=6\); \(n = 2\), \(6\cdot2^{1}=12\); \(n = 3\), \(6\cdot2^{2}=24

eq18\), so wrong.

• Option B: \(a_n=6\cdot2^{n}\). For \(n = 1\), \(6\cdot2^{1}=12

eq6\), wrong.

• Option D: \(a_n=6 + 6n\). For \(n = 1\), \(6+6\times1=12

eq6\), wrong.

• Option E: \(a_n=12-6n\). For \(n = 1\), \(12 - 6\times1=6\); \(n = 2\), \(12-12 = 0

eq12\) (assuming \(a_2\) should be \(12\) as per arithmetic sequence with \(d = 6\)), wrong.

Answer:

C. \(a_n = 6n\)