(b)The 9^th term of an arithmetic progression is twice as great as the 3^rd term, and the 15^th term is 27. Evaluate the sum of the first 25 terms of the series (3points) (c)The 4^th term of a geometric progression is 54 and the 7^th term is 1458. Calculate the sum of the first 10 terms (3points) (d)The 5^th and the 11^th terms of an arithmetic progression are 17 and 38 respectively. Determine the number sum of the first 20 terms of the series (3points) (e)Solve the following set of equations using the Gaussian elimination method in matrices x_(1)-4x_(2)-2x_(3)=21 2x_(1)+x_(2)+2x_(3)=3 3x_(1)+2x_(2)-x_(3)=-2 (5points) (f)Simplify the following complex numbers completely (1)int _(0)^7(1)int _(0)^7 (1point) and some of the desk (2points) (sin)_(d)^3 (1point) (w)]^37 (3points)

(b) To find the sum of the first 25 terms of the arithmetic progression, we need to determine the common difference and the first term.<br /><br />Given that the 9th term is twice as great as the 3rd term, we can write the equation:<br />\[ a_9 = 2a_3 \]<br />\[ a + 8d = 2(a + 2d) \]<br />\[ a + 8d = 2a + 4d \]<br />\[ 4d = a \]<br />\[ a = 4d \]<br /><br />Also, given that the 15th term is 27, we can write:<br />\[ a_{15} = a + 14d = 27 \]<br />\[ 4d + 14d = 27 \]<br />\[ 18d = 27 \]<br />\[ d = \frac{27}{18} = \frac{3}{2} \]<br /><br />Now, we can find the first term:<br />\[ a = 4d = 4 \times \frac{3}{2} = 6 \]<br /><br />The sum of the first 25 terms of the arithmetic progression is given by:<br />\[ S_{25} = \frac{n}{2} (2a + (n-1)d) \]<br />\[ S_{25} = \frac{25}{2} (2 \times 6 + 24 \times \frac{3}{2}) \]<br />\[ S_{25} = \frac{25}{2} (12 + 36) \]<br />\[ S_{25} = \frac{25}{2} \times 48 \]<br />\[ S_{25} = 600 \]<br /><br />(c) To find the sum of the first 10 terms of the geometric progression, we need to determine the common ratio and the first term.<br /><br />Given that the 4th term is 54 and the 7th term is 1458, we can write the equations:<br />\[ a_4 = ar^3 = 54 \]<br />\[ a_7 = ar^6 = 1458 \]<br /><br />Dividing the second equation by the first equation, we get:<br />\[ \frac{ar^6}{ar^3} = \frac{1458}{54} \]<br />\[ r^3 = 27 \]<br />\[ r = 3 \]<br /><br />Now, we can find the first term:<br />\[ a = \frac{54}{3^3} = \frac{54}{27} = 2 \]<br /><br />The sum of the first 10 terms of the geometric progression is given by:<br />\[ S_{10} = \frac{a(1-r^n)}{1-r} \]<br />\[ S_{10} = \frac{2(1-3^{10})}{1-3} \]<br />\[ S_{10} = \frac{2(1-59049)}{-2} \]<br />\[ S_{10} = \frac{2(-59048)}{-2} \]<br />\[ S_{10} = 29524 \]<br /><br />(d) To find the sum of the first 20 terms of the arithmetic progression, we need to determine the common difference and the first term.<br /><br />Given that the 5th term is 17 and the 11th term is 38, we can write the equations:<br />\[ a_5 = a + 4d = 17 \]<br />\[ a_{11} = a + 10d = 38 \]<br /><br />Subtracting the first equation from the second equation, we get:<br />\[ (a + 10d) - (a + 4d) = 38 - 17 \]<br />\[ 6d = 21 \]<br />\[ d = \frac{21}{6} = \frac{7}{2} \]<br /><br />Now, we can find the first term:<br />\[ a = 17 - 4d = 17 - 4 \times \frac{7}{2} = 17 - 14 = 3 \]<br /><br />The sum of the first 20 terms of the arithmetic progression is given by:<br />\[ S_{20} = \frac{n}{2} (2a + (n-1)d) \]<br />\[ S_{20} = \frac{20}{2} (2 \times 3 + 19 \times \frac{7}{2}) \]<br />\[ S_{20} = 10 (6 + \frac{133}{2}) \]<br />\[ S_{20} = 10 \times \frac{145}{2} \]<br />\[ S_{20} = 725 \]<br /><br />(e) To solve the set of equations using the Gaussian elimination method in matrices, we can write the augmented matrix:<br />\[ \begin{bmatrix} 1 & -4 & -2 & | & 21 \\ 2 & 1 & 2 & | & 3 \\ 3 & 2 & -1 & | & -2 \end{bmatrix} \]<br /><br />Perform