Given \( A_{t}=A_{0}(1+r t) \) Make \( t \) the subject of the formula. By referring to question 1.5 .1 above, calculate the value of \( t \) given that \( r=\frac{15}{100}, A_{t}=2625, A_{0}=1500 \). A brick layer wants to build a wall and he finds that he will need 25 bricks in the bottom row, 30 bricks in the second row and 35 bricks in the third row, up to the \( 20^{\text {th }} \) row. How many bricks are there in the last row? How many bricks are there in the wall? The sum of the first 15 terms of an arithmetic sequence is 165 . The common difference is 2 . Calculate the first term of the sequence.

To make \( t \) the subject of the formula \( A_{t} = A_{0} (1 + r t) \), we first divide both sides by \( A_{0} \) to get \( \frac{A_{t}}{A_{0}} = 1 + r t \). Next, we subtract 1 from both sides to yield \( \frac{A_{t}}{A_{0}} - 1 = r t \). Finally, we divide by \( r \) to isolate \( t \): \[ t = \frac{\frac{A_{t}}{A_{0}} - 1}{r} \] Now, substituting the values \( r = \frac{15}{100} = 0.15\), \( A_{t} = 2625 \), and \( A_{0} = 1500 \): \[ t = \frac{\frac{2625}{1500} - 1}{0.15} = \frac{1.75 - 1}{0.15} = \frac{0.75}{0.15} = 5 \] So, \( t \) equals 5. For the bricklayer's wall, the number of bricks in each row forms an arithmetic sequence where the first row has 25 bricks and the common difference is 5 (since the second row has 30 and the third row has 35). The number of bricks in the 20th row can be calculated as: \[ a_n = a + (n-1)d = 25 + (20 - 1)5 = 25 + 95 = 120 \] Thus, there are 120 bricks in the last (20th) row. To determine the total number of bricks in the wall, we can sum the first 20 terms of the arithmetic sequence: \[ S_n = \frac{n}{2} (a + a_n) = \frac{20}{2} (25 + 120) = 10 \times 145 = 1450 \] So, there are 1450 bricks in total in the wall. Now, for the arithmetic sequence whose sum is 165, with a common difference of 2: The sum of the first \( n \) terms is given by \( S_n = \frac{n}{2} (2a + (n-1)d) = 165 \). Plugging our known values into this equation, we get: \[ 165 = \frac{15}{2} (2a + (15-1)2) = \frac{15}{2} (2a + 28) \] Multiplying both sides by 2 to eliminate the fraction: \[ 330 = 15(2a + 28) \implies 22 = 2a + 28 \implies 2a = 22 - 28 \implies 2a = -6 \implies a = -3 \] Thus, the first term of the sequence is \(-3\).