Find the sixth and twentieth terms and the sum of the first 10term of each of the following Sequences [ -15,-9,-3, ldots.-9-(-15)=6 ]

To find the sixth and twentieth terms of the sequence, we can use the formula for the nth term of an arithmetic sequence:<br /><br />\[ a_n = a_1 + (n-1)d \]<br /><br />where \( a_n \) is the nth term, \( a_1 \) is the first term, \( n \) is the term number, and \( d \) is the common difference.<br /><br />In this case, the first term \( a_1 \) is -15, and the common difference \( d \) is 6 (since the difference between consecutive terms is 6).<br /><br />Let's find the sixth term:<br /><br />\[ a_6 = -15 + (6-1) \times 6 \]<br />\[ a_6 = -15 + 5 \times 6 \]<br />\[ a_6 = -15 + 30 \]<br />\[ a_6 = 15 \]<br /><br />So, the sixth term of the sequence is 15.<br /><br />Now, let's find the twentieth term:<br /><br />\[ a_{20} = -15 + (20-1) \times 6 \]<br />\[ a_{20} = -15 + 19 \times 6 \]<br />\[ a_{20} = -15 + 114 \]<br />\[ a_{20} = 99 \]<br /><br />So, the twentieth term of the sequence is 99.<br /><br />To find the sum of the first 10 terms of the sequence, we can use the formula for the sum of the first n terms of an arithmetic sequence:<br /><br />\[ S_n = \frac{n}{2}(2a_1 + (n-1)d) \]<br /><br />where \( S_n \) is the sum of the first n terms, \( n \) is the number of terms, \( a_1 \) is the first term, and \( d \) is the common difference.<br /><br />In this case, the first term \( a_1 \) is -15, the common difference \( d \) is 6, and we want to find the sum of the first 10 terms (\( n = 10 \)):<br /><br />\[ S_{10} = \frac{10}{2}(2(-15) + (10-1) \times 6) \]<br />\[ S_{10} = 5(-30 + 9 \times 6) \]<br />\[ S_{10} = 5(-30 + 54) \]<br />\[ S_{10} = 5 \times 24 \]<br />\[ S_{10} = 120 \]<br /><br />So, the sum of the first 10 terms of the sequence is 120.