Respuesta :

lastbenchstudent
this is a geometric progression

so first term a = 3

common ratio r = 12/3 or 48/12 = 4

so in GP n term is given as

[tex] t_{n} \: = a {r}^{n - 1} \\ so \: \\ t_{1} \: = a {r}^{1 - 1} = 3 \times {r}^{0} = 3 \times 1 = 3 \\ t_{2} \: = a {r}^{2 - 1} = 3 \times {r}^{1} = 3 \times {4}^{1} = 3 \times 4 = 12 \\ t_{15} \: = a {r}^{15 - 1} = 3 \times {r}^{14} = 3 \times {4}^{14} = 3 \times 268435456 = 805306368 \\ [/tex]

so answer is 805306368
zainebamir540

Answer:

a₁₅ =  805306368 is the 15th term of the sequence.

Step-by-step explanation:

We have given a sequence :

3,12,48,192,768,...............

TO find is it geometric sequence or not:

12/3 = 4

48/12 = 4

The ratio is same so it is a geometric sequence.

common ratio= r = 4

The formula to find the n terms is given  by:

aₙ =  arⁿ⁻¹

First term = a = 3 , r = 4.

We have to find the 15th term of the sequence.

put n = 15 we get,

a₁₅ = 3(4)¹⁵⁻¹

a₁₅ = 3(4)¹⁴ = 3 × 268435456= 805306368.

a₁₅ =  805306368 is the 15th term of the sequence.

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