Find the sum of the geometric sequence;
4/3,8/3,16/3,32/3,64/3

a.) 122/3
b.)124/3
c.)124/15
d.)122/5

you for your answer!

124/3

Step-by-step explanation:

The first term, a is 4/3

The common ratio, r is 2

The number of terms, n is 5

Therefore;

Sum = a( r^n-1)/(r-1)

        = 4/3 ((2^5) -1)/(2-1)

        = 4/3 × ((2^5) -1)

        = 4/3 × (32-1)

         = 4/3× 31

         = 124/3

step-by-step explanation:

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