Respuesta :

Poltergeist

Answer:

[tex]a_1=12[/tex]

Step-by-step explanation:

Assuming the geometric series is

[tex]$\sum_{n=1}^{\infty} 12*(-\frac{1}{3} )^{n-1}$[/tex]

then nth term is

[tex]a(n)=12*(-\frac{1}{3} )^{n-1}[/tex],

and the first term is found when [tex]n=1[/tex]:

[tex]a_1=12*(-\frac{1}{3} )^{1-1}=12*(-\frac{1}{3} )^0\\\\\boxed{a_1=12}[/tex]

abidemiokin

Answer:

a1 = 12

Step-by-step explanation:

Given the geometric series

Sigma-Summation Underscript n = 1 Overscript infinity EndScripts 12 (negative one-ninth) Superscript n minus 1

From the series given, the nth term of the sequence is given as;

an = 12(1/9)^n-1

To get a1, we will substitute the value of n=1 into the nth term of Tue geometric series.

Given the nth term of the series as;

an = 12(1/9)^n-1

When n=1

a1 = 12(1/9)^1-1

a1 = 12(1/9)^0

Since anything raise to power of zero is 1, then;

(1/9)^0 = 1

a1 = 12{1}

a1 = 12

The value of a1 of the geometric series is 12.

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