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Claire saves $\$45$ one month and then each month thereafter she saves $\$4$ more than the preceding month for a $16$ -year period. (a) What will her savings be during the last month of the $16$ -year period?

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Identify statistical questions

Full solution

Q. Claire saves

\$45

one month and then each month thereafter she saves

\$4

more than the preceding month for a

16

-year period. (a) What will her savings be during the last month of the

16

-year period?

Calculate Total Months: Determine the total number of months in a $16$ -year period. $\newline$ There are $12$ months in a year, so in $16$ years, there will be $16 \times 12$ months. $\newline$ $16 \times 12 = 192$ months
Identify Savings Pattern: Identify the pattern of savings increase each month. Claire saves $\$45$ in the first month and then each month thereafter she saves $\$4$ more than the preceding month. This is an arithmetic sequence with the first term $a_1 = \$45$ and a common difference $d = \$4$ .
Calculate Last Month Savings: Calculate the savings for the last month. $\newline$ The savings in the last month will be the $192^{\text{nd}}$ term of the arithmetic sequence. The $n^{\text{th}}$ term of an arithmetic sequence is given by $a_n = a_1 + (n - 1)d$ . $\newline$ Substitute $a_1 = \$45$ , $d = \$4$ , and $n = 192$ to find the $192^{\text{nd}}$ term. $\newline$ $a_{192} = \$45 + (192 - 1) \times \$4$
Perform Calculation: Perform the calculation for the $192$ nd term. $\newline$ $a_{192} = \$45 + 191 \times \$4$ $\newline$ $a_{192} = \$45 + \$764$ $\newline$ $a_{192} = \$809$

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