Math, asked by Singh98gss, 1 year ago

Sum of infinite series 1-1/3+1/9-1/27+.....

Answers

Answered by kartavyaguptalm
0

Answer:

The sum of the given infinite series is found to be 0.75 or \frac{3}{4}.

Step-by-step explanation:

The given infinite series is as follows:

1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+.....

On carefully observing, we can say that the given infinite series is a geometric series with a common ratio.

Finding the common ratio, we get:

\frac{a_2}{a_1}=\frac{-\frac{1}{3}}{1}=-\frac{1}{3}

similarly,

\frac{a_3}{a_2}=\frac{\frac{1}{9}}{-\frac{1}{3}}=-\frac{1}{3}

Thus, the common ratio of the given series is -\frac{1}{3}.

Now, we know that the expression of the sum of infinite series is given by:

S_\infty=\frac{a}{1-r}

Substituting the known information, we get:

S_\infty=\frac{1}{1-(-\frac{1}{3})}

or we can say:

S_\infty=\frac{1}{1+\frac{1}{3}}

S_\infty=\frac{1}{\frac{1+3}{3}}

which gives us:

S_\infty=\frac{3}{4} or 0.75

Thus, the sum of the given infinite series is found to be 0.75 or \frac{3}{4}.

Answered by rani78956
0

In Maths, Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio. This progression is also known as a geometric sequence of numbers that follow a pattern.

Given:

1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+...

a=1, a_2=\frac{-1}{3}, a_3=\frac{1}{9}, a_4=\frac{-1}{27}

r=\frac{a_2}{a_1}=\frac{\frac{-1}{3} }{1}=\frac{-1}{3}

Sum of infinite terms of G.P=\frac{a}{1-r}

=\frac{1}{1-[\frac{-1}{3} ]}\\

=\frac{1}{1-[\frac{-1}{3}] }

=\frac{1}{\frac{4}{3} }

=\frac{3}{4}.

Similar questions