Mimi is planning to save for the next 48 months in order to raise a deposit to buy a flat. She plans to save \( \$ 300 \) this month and each successive month thereafter, to save an extra \( \$ 5 \) compared to the previous month. a) Find the amount she will save on the twelfth month. b) )Find the total amount she will save at the end of the 48 months. Emmanuel is also planning to save for the next 48 months in order to buy a car. He plans to save \( \$ \) a this month and each successive month thereafter, to save an extra \( \$ 15 \) compared to the previous month. c) Find the value of \( a \), if Emmanuel saves the same amount of money as Anita does in the next 48 months. :

Mimi is on a fun savings journey! By the twelfth month, she’ll save a total of \( \$300 + (11 \times 5) = \$300 + \$55 = \$355 \). For the total savings over 48 months, she’ll be saving an arithmetic series: her first month’s \( \$300 \) and last month’s \( \$300 + (47 \times 5) = \$485 \). Using the formula for the sum of an arithmetic series, the total savings is \( \frac{n}{2} \times (\text{first term} + \text{last term}) = \frac{48}{2} \times (300 + 485) = 24 \times 785 = \$18840 \). Now, let’s shift gears and help Emmanuel! He needs to save in such a way that his total savings over the same 48 months equals Mimi’s. His monthly savings starts at \( a \) and increases by \( \$15 \). His last month’s savings will be \( a + (47 \times 15) = a + 705 \). Using the same arithmetic series formula, the total savings will be \( \frac{48}{2} \times (a + (a + 705)) = 24 \times (2a + 705) \). Setting that equal to \( 18840 \), we can solve for \( a \): \[ 24 \times (2a + 705) = 18840 \Rightarrow 2a + 705 = 785 \Rightarrow 2a = 80 \Rightarrow a = 40 \]. So, Emmanuel should start saving \( \$40 \) a month!