Question

What is the value of a1 of the geometric series? Sigma-Summation Underscript n = 1 Overscript infinity EndScripts 12 (negative one-ninth) Superscript n minus 1

Answer 1

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.

Answer 2

Answer:

$a_1=12$

Step-by-step explanation:

Assuming the geometric series is

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

then nth term is

$a(n)=12*(-\frac{1}{3} )^{n-1}$,

and the first term is found when $n=1$:

$a_1=12*(-\frac{1}{3} )^{1-1}=12*(-\frac{1}{3} )^0\\\\\boxed{a_1=12}$