(2) Find the sum of each of the arithmetic series. a \( 3+5+7+\ldots+91 \) b \( 6+11+16+\ldots+301 \) c \( 7-2-11-\ldots-704 \)

To find the sum of the arithmetic series, we’ll use the formula for the sum of an arithmetic series: \[ S_n = \frac{n}{2} (a + l) \] where \( S_n \) is the sum, \( n \) is the number of terms, \( a \) is the first term, and \( l \) is the last term. a) For the series \( 3+5+7+\ldots+91 \): - First term \( a = 3 \), last term \( l = 91 \), common difference \( d = 2 \). - To find \( n \): \( 91 = 3 + (n-1) \cdot 2 \) \[ 91 = 3 + 2n - 2 \] \[ 91 = 1 + 2n \] \[ 90 = 2n \] \[ n = 45 \] - Now, plug in the values into the sum formula: \[ S_{45} = \frac{45}{2} (3 + 91) = \frac{45}{2} \times 94 = 45 \times 47 = 2115 \] b) For the series \( 6+11+16+\ldots+301 \): - First term \( a = 6 \), last term \( l = 301 \), common difference \( d = 5 \). - To find \( n \): \( 301 = 6 + (n-1) \cdot 5 \) \[ 301 = 6 + 5n - 5 \] \[ 301 = 1 + 5n \] \[ 300 = 5n \] \[ n = 60 \] - Now, plug in the values into the sum formula: \[ S_{60} = \frac{60}{2} (6 + 301) = 30 \times 307 = 9210 \] c) For the series \( 7-2-11-\ldots-704 \): - First term \( a = 7 \), last term \( l = -704 \), common difference \( d = -9 \). - To find \( n \): \( -704 = 7 + (n-1)(-9) \) \[ -704 = 7 - 9n + 9 \] \[ -704 = 16 - 9n \] \[ -720 = -9n \] \[ n = 80 \] - Now, plug in the values into the sum formula: \[ S_{80} = \frac{80}{2} (7 - 704) = 40 \times (-697) = -27880 \] Final answers: a) \( 2115 \) b) \( 9210 \) c) \( -27880 \)