Victory for the ACT Student Text 15e

218 • M ATH

EVALUATING SEQUENCES

15. In a geometric sequence of positive numbers, the fourth term is 125 and the sixth term is 3,125. What is the second term of the sequence? A. 1 B. 5 C. 10 D. 25 E. 50 City University projects that a planned expansion will increase the number of enrolled students every year for the next ϐ‹˜‡ ›‡ƒ”• „› ͷͲΨǤ ˆ ͶͲͲ •–—†‡–• ‡”‘ŽŽ ‹ –Š‡ ϐ‹”•– ›‡ƒ” ‘ˆ –Š‡ ’Žƒǡ Š‘™ ƒ› •–—†‡–• ƒ”‡ ‡š’‡…–‡† –‘ ‡”‘ŽŽ ‹ –Š‡ ϐ‹ˆ–Š year of the plan? F. 25 G. 600 H. 675 J. 1,350 K. 2,025 Jimmy’s uncle deposited $1,000 into a college fund account and promised that at the start of each year, he would †‡’‘•‹– ƒ ƒ‘—– ‡“—ƒŽ –‘ ͳͲΨ ‘ˆ –Š‡ account balance. If no other deposits or withdrawals were made and no additional interest accrued, what was the account balance after three additional annual deposits were made by Jimmy’s uncle? A. $1,030 B. $1,300 C. $1,331 D. $1,500 E. $1,830 16. 17.

A tank with a capacity of 2,400 liters is ϐ‹ŽŽ‡† –‘…ƒ’ƒ…‹–› ™‹–Š ™ƒ–‡”Ǥ ˆ ƒ ˜ƒŽ˜‡ ‹• ‘’‡‡† –Šƒ– †”ƒ‹• ʹͷΨ ‘ˆ –Š‡…‘–‡–• ‘ˆ the tank every minute, what is the volume of water (in liters) that remains in the tank after 3 minutes? F. 1,800 G. 1,350 H. 1,012.5 J. 600 K. 325.75 Throwback to Math Class Exponential growth and decay are special cases of geometric sequences, with the formula a a r t T t 0 . In the above equation, a t is the amount after time t , a 0 is the initial amount ( t 0 ), r is the proportionality constant, t is the total period of growth (decay), and T is the time per cycle of constant growth (decay). Note that for exponential growth, r 1 ! ; for exponential decay, r 0 1 ; and if r 1 , there is no change in amount.

18.

Throwback to Math Class In exponential growth sequences, each term is a constant multiple of the preceding term. For example, consider the geometric sequence 3, 9, 27, 81,…. Each consecutive term is equal to 3 times the preceding term, so the common ratio between terms is 3. Here’s a more ˆ‘”ƒŽ †‡ϐ‹‹–‹‘ǣ Any term ( a n ) is related to the preceding term ( ) a n 1 by the common ratio ( r ): a a r n n 1 = − The relationship between the n th term ( a n ) and the ϐ‹”•– –‡” ȋ a 1 ) is as follows: a a r n n 1 1 = −