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### Geometric Sequence Phrase Issues

This lesson will present you how you can remedy a wide range of geometric sequence phrase issues.

Instance #1:

The inventory’s worth of an organization shouldn’t be doing properly currently. Suppose the inventory’s worth is 92% of its earlier worth every day. What’s the inventory’s worth after 10 days if the inventory was value \$2500 proper earlier than it began to go down?

To resolve this drawback, we’d like the geometric sequence components proven under.

an = a1 × r(n – 1)

a1 = authentic worth of the inventory  = 2500

a2 = worth of the inventory after 1 day

a11 = worth of the inventory after 10 days

r = 0.92

a11 = 2500 × (0.92)(11 – 1)

a11 = 2500 × (0.92)10

a11 = 2500 × 0.434

a11 = \$1085

The inventory’s worth is about 1085 {dollars}.

Instance #2:

The third time period of a geometrical sequence is 45 and the fifth time period of the geometric sequence is 405. If all of the phrases of the sequence are optimistic numbers, discover the fifteenth time period of the geometric sequence.

To resolve this drawback, we’d like the geometric sequence components proven under.

an = a× r(n – 1)

Discover the third time period

a3 = a1 × r(3 – 1)

a3 = a1 × r2

Because the third time period is 45,  45 = a1 × r2 (equation 1)

Discover the fifth time period

a5 = a1 × r(5 – 1)

a5 = a1 × r4

Because the fifth time period is 405,  405 = a1 × r4 (equation 2)

Divide equation 2 by equation 1.

(a1 × r4) / (a1 × r2) = 405 / 45

Cancel a1 since it’s each on prime and on the backside of the fraction.

r4 / r2 = 9

r2 = 9

r = ±√9

r = ±3

Use r  = 3, and equation 1 to discover a1

45 = a1 × (3)2

45 = a1 × 9

a1 = 45 / 9 = 5

Since all of the phrases of the sequence are optimistic numbers, we should use r = 3 if we would like all of the phrases to be optimistic numbers.

an = a1 × r(n – 1)

Allow us to now discover a15

a15 = 5 × (3)(15 – 1)

a15 = 5 × (3)14

a15 = 5 × 4782969

a15 =  23914845

## Difficult geometric sequence phrase issues

Instance #3:

Suppose that the magnification of a PDF file on a desktop laptop is elevated by 15% for every degree of zoom. Suppose additionally that the unique size of the phrase “January” is 1.2 cm. Discover the size of the phrase “January” after 6 magnifications.

To resolve this drawback, we’d like the geometric sequence components proven under.

an = a1 × r(n – 1)

a1 = authentic size of the phrase  = 1.2 cm

a2 = size of the phrase after 1 magnification

a7 = size of the phrase after 6 magnifications

r = 1 + 0.15 = 1.15

n = 7

a7 = 1.2 × (1.15)(7 – 1)

a7 = 1.2 × (1.15)6

a7 = 1.2 × (1.15)6

a7 = 1.2 × 2.313

a7 = 2.7756

After 6 magnifications, the size of the phrase “January” is 2.7756 cm.

Discover that we added 1 to 0.15. Why did we do this? Allow us to not use the components immediately so you possibly can see the rationale behind it. Examine the next fastidiously!

Day 1: a1 = 1.2

Day 2: a2 = 1.2 + 1.2(0.15) = 1.2(1 + 0.15)

Day 3: a31.2(1 + 0.15) + [1.2(1 + 0.15)]0.15 = 1.2(1 + 0.15)(1 + 0.15) = 1.2(1 + 0.15)2

Day 7: a7 = 1.2(1 + 0.15)6

Instance #4

Suppose that you really want a decreased copy of {a photograph}. The precise size of the {photograph} is 10 inches. If every discount is 64% of the unique, what number of reductions, will shrink the {photograph} to 1.07 inches.

an = a1 × r(n – 1)

a1 = authentic size of the {photograph}  = 10 inches

a2 = size of the {photograph} after 1 discount

an = 1.07

r = 0.64

n = variety of reductions = ?

1.07 = 10 × (0.64)(n – 1)

Divide either side by 10

1.07 / 10 = [10 × (0.64)(n – 1)] / 10

0.107 = (0.64)(n – 1)

Take the pure log of either side of the equation.

ln(0.107) = ln[(0.64)(n – 1)]

Use the energy property of logarithms.

ln(0.107) = (n – 1)ln(0.64)

Divide either side of the equation by ln(0.64)

ln(0.107) / ln(0.64) = (n – 1)ln(0.64) / ln(0.64)

n – 1 = ln(0.107) / ln(0.64)

Use a calculator to search out ln(0.107) and ln(0.64)

n – 1 = -2.23492644452 -0.44628710262

n – 1 = 5.0078

n = 1 + 5.0078

n = 6.0078

Subsequently, you will have 6 reductions.