Question

Find the number of term of the series 1/27,1/9/,1/3,....729

Answer 1

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•Find the number of term of the series 1/27,1/9/,1/3,....729.

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$first \: term(a) = \frac{1}{27} \\common \: ratio( r) = \frac{ \frac{1}{9} }{ \frac{1}{27} } = 3$

let the n'th term is =729

now...

using the formula..

$t(n) = ar {}^{n - 1}$

here...t(n)=729

r=3

$a = \frac{1}{27}$

therefore....

$729 = \frac{1}{27} \times 3 {}^{n - 1} \\ = > 729 \times 27 = 3 {}^{n - 1} \\ = > 3 {}^{6} \times 3 {}^{3} = 3 {}^{n - 1} \\ = > 3 {}^{9} = 3 {}^{n - 1} \\ > n - 1 = 9 \\ = > n = 10$

so the 10th term of the series is =729

$<font color=green><marquee direction="Right">hope this help you$

Answer 2

For the given series $1/27,1/9/,1/3,....729$ the number of terms is 10

$(r)=\frac{\frac{1}{9}}{\frac{1}{27}}=3$

Geometric series

an infinite series of the form a + ar + ar2 + ar3+⋯, where r is known as the common ratio. A simple example is the geometric series for a = 1 and r = 1/2, or 1 + 1/2 + 1/4 + 1/8 +⋯,

concept to use for calculation of nth term of series

geometric sequence with first term a and the common ratio r is given by :

$a_{n}=t_{n}=a r^{n-1}$ where a be the first term and r be the common ratio for a given Geometric Sequence.

Given :

First term $(a) =1/27$

common ratio $(r)=\frac{\frac{1}{9}}{\frac{1}{27}}=3$

Step by step solution

Calculation of the numbers of term for the series

The n'th term is$=729$

$t(n)=a r^{n-1}$

$t (n)=729$

$r=3$

$a=\frac{1}{27}$

Substitute the given values in above formula

$729=\frac{1}{27} \times 3^{n-1}$

$= > 729 \times 27=3^{n-1}$

$= > 3^{6} \times 3^{3}=3^{n-1}$

$= > 3^{9}=3^{n-1}$

$> n-1=9$

$= > n=10$

Hence the numbers of term in the series is 10

Learn more about geometric series here,

https://brainly.in/textbook-solutions/q-answer-following-questions-define-geometric-series

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