Consider the sequences. Use the equation to determine the \( 10^{\text {th }} \) term and the nth term. a) \( 10 ; 16 ; 22 ; 28 \) b) \( 33 ; 30 ; 27 ; 24 \) c) \( 5 ;-1 ;-7 \) d) \( -4 ;-11 ;-18:-25 \)
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For the first sequence \(10, 16, 22, 28\), we can see it's an arithmetic sequence with a common difference of 6. To find the \(n^{\text{th}}\) term, we can use the formula \(a_n = a_1 + (n-1)d\), where \(a_1 = 10\) and \(d = 6\). Thus, \(a_n = 10 + (n-1)6 = 6n + 4\). Plugging in \(n=10\), we get \(a_{10} = 6(10) + 4 = 64\). For the second sequence \(33, 30, 27, 24\), it's also an arithmetic progression but with a negative common difference of 3. Using the same formula, \(a_1 = 33\) and \(d = -3\), we get \(a_n = 33 + (n-1)(-3) = -3n + 36\). Thus, \(a_{10} = -3(10) + 36 = 6\). Now, moving on to the third sequence \(5, -1, -7\), this one has a common difference of -6. Here, \(a_1 = 5\) and \(d = -6\), so \(a_n = 5 + (n-1)(-6) = -6n + 11\). The \(n^{\text{th}}\) term is found with \(n=10\): \(a_{10} = -6(10) + 11 = -49\). Finally, for the last sequence \(-4, -11, -18, -25\), this is yet another arithmetic sequence with a common difference of -7. Here, \(a_1 = -4\) and \(d = -7\), giving us \(a_n = -4 + (n-1)(-7) = -7n + 3\). So plugging in \(n=10\), \(a_{10} = -7(10) + 3 = -67\). To summarize: 1. For \(10, 16, 22, 28\), \(10^{\text{th}}\) term is \(64\), \(n^{\text{th}}\) term is \(6n + 4\). 2. For \(33, 30, 27, 24\), \(10^{\text{th}}\) term is \(6\), \(n^{\text{th}}\) term is \(-3n + 36\). 3. For \(5, -1, -7\), \(10^{\text{th}}\) term is \(-49\), \(n^{\text{th}}\) term is \(-6n + 11\). 4. For \(-4, -11, -18, -25\), \(10^{\text{th}}\) term is \(-67\), \(n^{\text{th}}\) term is \(-7n + 3\).
