1.3 Consider the geometric series: \( 5+10+20+\cdots \) 1.3.1 How many terms of the series add up to 20475 ? 1.3.2 Determine the least of terms for which the sum of the series will be greater than 5000 .

To solve this geometric series, we identify the first term \( a = 5 \) and the common ratio \( r = 2 \). The formula for the sum \( S_n \) of the first \( n \) terms of a geometric series is given by \( S_n = a \frac{r^n - 1}{r - 1} \). To find how many terms add up to 20475, set \( S_n = 20475 \): \[ 5 \frac{2^n - 1}{2 - 1} = 20475 \] This simplifies to: \[ 5(2^n - 1) = 20475 \] \[ 2^n - 1 = \frac{20475}{5} = 4095 \] \[ 2^n = 4096 \] Given that \( 2^{12} = 4096 \), we get \( n = 12 \). Next, to determine the least number of terms needed for the sum to exceed 5000, we set up the inequality: \[ 5 \frac{2^n - 1}{2 - 1} > 5000 \] This simplifies to: \[ 5(2^n - 1) > 5000 \] \[ 2^n - 1 > 1000 \] \[ 2^n > 1001 \] Calculating \( 2^{10} = 1024 \) and \( 2^9 = 512 \), we find \( n = 10 \) is the smallest integer satisfying this inequality. In summary, 12 terms sum up to 20475, and at least 10 terms are required for the sum to exceed 5000.