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Claire saves $45\$45 one month and then each month thereafter she saves $4\$4 more than the preceding month for a 1616-year period. (a) What will her savings be during the last month of the 1616-year period?

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Q. Claire saves $45\$45 one month and then each month thereafter she saves $4\$4 more than the preceding month for a 1616-year period. (a) What will her savings be during the last month of the 1616-year period?
  1. Calculate Total Months: Determine the total number of months in a 1616-year period.\newlineThere are 1212 months in a year, so in 1616 years, there will be 16×1216 \times 12 months.\newline16×12=19216 \times 12 = 192 months
  2. Identify Savings Pattern: Identify the pattern of savings increase each month. Claire saves $45\$45 in the first month and then each month thereafter she saves $4\$4 more than the preceding month. This is an arithmetic sequence with the first term a1=$45a_1 = \$45 and a common difference d=$4d = \$4.
  3. Calculate Last Month Savings: Calculate the savings for the last month.\newlineThe savings in the last month will be the 192nd192^{\text{nd}} term of the arithmetic sequence. The nthn^{\text{th}} term of an arithmetic sequence is given by an=a1+(n1)da_n = a_1 + (n - 1)d.\newlineSubstitute a1=$45a_1 = \$45, d=$4d = \$4, and n=192n = 192 to find the 192nd192^{\text{nd}} term.\newlinea192=$45+(1921)×$4a_{192} = \$45 + (192 - 1) \times \$4
  4. Perform Calculation: Perform the calculation for the 192192nd term.\newlinea192=$45+191×$4a_{192} = \$45 + 191 \times \$4\newlinea192=$45+$764a_{192} = \$45 + \$764\newlinea192=$809a_{192} = \$809

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