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The population of a type of local bass can be found using an infinite geometric series where a1 = 94 and the common ratio is one third. Find the sum of this infinite series that will be the upper limit of this population.

Verified Answer :

a/(1-r)


a = 94

r = 1/3

94/(1-1/3)

= 141

Answer:

The sum of this infinite series that will be the upper limit of this population is, 141

Step-by-step explanation:

Formula for infinite  geometric series is given by:

\(S = \frac{a_1}{1-r}\)        ....[1]

where,

\(a_1\) is the first term,

r is the common ratio term.

As per the statement:

The population of a type of local bass can be found using an infinite geometric series where:

\(a_1 = 94\)

\(r = \frac{1}{3}\)

To find the sum of this infinite series that will be the upper limit of this population.

Substitute the given values in [1] we have;

\(S = \frac{94}{1-\frac{1}{3}} = \frac{94}{\frac{2}{3}} = 94 \cdot \frac{3}{2} = 47 \cdot 3 = 141\)

Therefore, 141 is the sum of this infinite series that will be the upper limit of this population.