3) Determine the no of term a, \( 3+11+19+\ldots \ldots=1580 \)

To determine the number of terms in the arithmetic series \( 3 + 11 + 19 + \ldots = 1580 \), we can follow these steps: ### Step 1: Identify the Given Information - **First term (\(a\))**: 3 - **Common difference (\(d\))**: \(11 - 3 = 8\) - **Sum of the series (\(S_n\))**: 1580 ### Step 2: Use the Formula for the Sum of an Arithmetic Series The sum of the first \(n\) terms of an arithmetic series is given by: \[ S_n = \frac{n}{2} \times [2a + (n - 1)d] \] Plugging in the known values: \[ 1580 = \frac{n}{2} \times [2 \times 3 + (n - 1) \times 8] \] \[ 1580 = \frac{n}{2} \times [6 + 8n - 8] \] \[ 1580 = \frac{n}{2} \times (8n - 2) \] \[ 1580 = \frac{8n^2 - 2n}{2} \] \[ 1580 = 4n^2 - n \] ### Step 3: Rearrange into a Quadratic Equation \[ 4n^2 - n - 1580 = 0 \] ### Step 4: Solve the Quadratic Equation Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -1 \), and \( c = -1580 \): \[ \Delta = b^2 - 4ac = (-1)^2 - 4 \times 4 \times (-1580) = 1 + 25280 = 25281 \] \[ \sqrt{25281} = 159 \] \[ n = \frac{1 \pm 159}{8} \] We discard the negative solution as the number of terms cannot be negative: \[ n = \frac{160}{8} = 20 \] ### Conclusion **There are 20 terms in the series.**