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Find the sum of the geometric sequence. 4 divided by 3, 8 divided by 3, 16 divided by 3, 32 divided by 3, 64 divided by 3

Respuesta :

swiftor
So this problem is just adding the dividends so it would be 4/3+8/3+16/3+32/3+64/3 so it would be 124/3 or 41.33333

Answer:

41.333

Step-by-step explanation:

Given is a geometric sequence

[tex]\frac{4}{3} ,\frac{8}{3} ,\frac{16}{3} ,\frac{32}{3} ,\frac{64}{3}[/tex]

No of terms n =5

a = I term =[tex]\frac{4}{3}[/tex]

Use the formula for sum of geometric sequence when r >1

[tex]S_n =\frac{a(r^n-1}{r-1} \\=\frac{4}{3}[\frac{ (2^5-1}{ 2 -1}] \\=4(\frac{31}{3} )\\=41.3333[/tex]

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