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1.

Sam puts \$1.25 into his piggy bank in January 2013. Each month he increases how much he puts in by 4 cents.

How much money will he put in his piggy bank in April 2016?

A)\$157.25

B)\$16.85

C)\$3.35

D)\$2.81                                                                                                                                     2.

Sarah's yearly salary is based on a geometric sequence. Her yearly salaries for her first three years are \$34,000 , \$35,020, and \$36,070.60

What will be her yearly salary for her 7th year?

A)\$41,140

B)\$40,000

C)\$40,597.78

D)\$41,815.71

Guest Dec 14, 2017
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#1
+1

1.

There are 39 months between January 2013 and April 2016.

39 x 0.04 cents =\$1.56 - the increase in his saving by April 2016

\$1.56 + \$1.25 =\$2.81 - Amount he must put in his piggy bank in April 2016.

2.

\$34,000 x 1.03^6 =\$40,597.78 - Sarah's expected salary by  7th year.

Or, you could use the nth term formula to find the same thing:

Nth. Term = ​F x R^(N - 1)

7th. term =\$34,000 x 1.03^6 =\$40,597.78 - Which is the same as above.

Guest Dec 14, 2017
#2
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2.   Find the common ratio  =

35020 / 34000  =   1.03

Salary for nth  year   =   34000 (1.03)n-1

Salary for the 7th year  =

34000(1/03)7-1  =  34000(1.03)6  =  \$40597.78

CPhill  Dec 14, 2017

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