Determine \( n \) in each of the arithmetic series: a) \( 16+19+22+\ldots \) (to \( n \) terms) \( =295 \) b) \( 6+17+28+\ldots \) (to \( n \) terms) \( =273 \)

To find \( n \) in each of the arithmetic series, we can use the formula for the sum of an arithmetic series: \[ S_n = \frac{n}{2} (a + l) \] where \( S_n \) is the sum of the first \( n \) terms, \( a \) is the first term, and \( l \) is the last term. We can also express \( l \) as \( a + (n-1)d \), where \( d \) is the common difference. ### a) For the series \( 16 + 19 + 22 + \ldots \) 1. First, identify the first term and common difference: - The first term \( a = 16 \) - The common difference \( d = 19 - 16 = 3 \) 2. We need to express the last term \( l \): \[ l = a + (n-1)d = 16 + (n-1) \cdot 3 \] 3. Substitute into the sum formula: \[ S_n = \frac{n}{2} (16 + l) = \frac{n}{2} \left(16 + 16 + (n-1) \cdot 3\right) = \frac{n}{2} \left(32 + (n-1) \cdot 3\right) \] Set this equal to 295: \[ \frac{n}{2} \left(32 + (n-1) \cdot 3\right) = 295 \] 4. Solve for \( n \): \[ n(32 + (n-1) \cdot 3) = 590 \] \[ n(32 + 3n - 3) = 590 \] \[ n(3n + 29) = 590 \] \[ 3n^2 + 29n - 590 = 0 \] 5. Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ n = \frac{-29 \pm \sqrt{29^2 - 4 \cdot 3 \cdot (-590)}}{2 \cdot 3} \] \[ n = \frac{-29 \pm \sqrt{841 + 7080}}{6} \] \[ n = \frac{-29 \pm \sqrt{7921}}{6} \] \[ n = \frac{-29 \pm 89}{6} \] This gives: \[ n = \frac{60}{6} = 10 \quad \text{(valid solution)} \] \[ n = \frac{-118}{6} \quad \text{(not valid)} \] ### b) For the series \( 6 + 17 + 28 + \ldots \) 1. Identify the first term and common difference: - The first term \( a = 6 \) - The common difference \( d = 17 - 6 = 11 \) 2. Express the last term \( l \): \[ l = a + (n-1)d = 6 + (n-1) \cdot 11 \] 3. Substitute into the sum formula: \[ S_n = \frac{n}{2} (6 + l) = \frac{n}{2} \left(6 + 6 + (n-1) \cdot 11\right) = \frac{n}{2} \left(12 + (n-1) \cdot 11\right) \] Set this equal to 273: \[ \frac{n}{2} (12 + (n-1) \cdot 11) = 273 \] 4. Solve for \( n \): \[ n(12 + (n-1) \cdot 11) = 546 \] \[ n(12 + 11n - 11) = 546 \] \[ n(11n + 1) = 546 \] \[ 11n^2 + n - 546 = 0 \] 5. Using the quadratic formula: \[ n = \frac{-1 \pm \sqrt{1^2 - 4