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

a1 = 12

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

`a_1=12`

Explanation:

Assuming the geometric series is

`$\sum_(n=1)^(\infty) 12*(-(1)/(3) )^(n-1)$`

then nth term is

`a(n)=12*(-(1)/(3) )^(n-1)`,

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

`a_1=12*(-(1)/(3) )^(1-1)=12*(-(1)/(3) )^0\\\\\boxed{a_1=12}`