how to find r in geometric series

A set of things that are in order is called a Sequence and when Sequences start to follow a certain pattern, they are known equally Progressions. Progressions are of dissimilar types like Arithmetics Progression, Geometric Progressions, Harmonic Progressions.

The sum of a particular Sequence is called a Series. A Series can be Space or Finite depending upon the Sequence, If a Sequence is Infinite, it will give Space Series whereas, if a Sequence is finite, information technology will give Finite serial.

Let's accept a finite Sequence:

a_i,a_ii,a_iii,a_4,a_{five,……….}a_n

The Series of this Sequence is given as:

a₁+ a_two+ a_iii+ a₄+a₅+……….a_north

The Series is too denoted as :

$\sum_{k=1}^{n}a_k$

Series is represented using Sigma (∑) Note in gild to Betoken Summation.

Geometric Series

In a Geometric Series, every next term is the multiplication of its Previous term by a certain constant and depending upon the value of the abiding, the Series may be Increasing or decreasing.

Geometric Sequence is given as:

a, ar, ar², ar³, ar⁴,….. {Infinite Sequence}

a, ar, ar², ar³, ar⁴, …
230;. arⁿ {Finite Sequence}
Geometric Series for the above is written equally:

a + ar + ar^two + ar³ + ar^iv +…. {Infinite Series}

a + ar + ar^two + ar^three + ar^iv +….. arⁿ {Finite Series}

Where. a = First term

r = Common Factor

Can the values of 'a' and 'r' be 0?

Answer: No, the value of a≠0, if the get-go term becomes goose egg, the series will not go along. Similarly, r≠0.

Geometric Series Formula

The Geometric Series formula for the Finite series is given as,

${S_n =\frac{a(1-r^n)}{1-r}}$

where, S = sum up to northward^thursday term

a = First term

r = common cistron

Derivation for Geometric Series Formula

Suppose a Geometric Series for n terms:

South_northward = a + ar + ar^two + ar^three + …. + ar^n-ane ⇢ (1)

Multiplying both sides past the mutual gene (r):

r S_north = ar + ar² + ar³ + ar⁴ + … + arⁿ ⇢ (2)

Subtracting Equation (one) from Equation (2):

(r S_n – S_n) = (ar + ar² + arⁱⁱⁱ + ar⁴ + … arⁿ) – (a + ar + ar² + ar³ +… + ar^north-1)

S_north (r-i) = ar^{due north} – a

South_n
b> (1 – r) = a (1-rⁿ)
${S_n =\frac{a(1-r^n)}{1-r}}$

Note: When the value of k starts from 'thou', the formula will change.

$\sum_{k=m}^{n}ar^k=\frac{a(r^m-r^{n+1}}{1-r}$ , when r≠0

For Infinite Geometric Serial

due north will tend to Infinity, n⇢∞, Putting this in the generalized formula:

$S_\infty = \sum_{northward=1}^{\infty}ar^{n-1} = \frac{a}{1-r}; -1<{r}<1$

Northward^th term for the Chiliad.P. : a_n = ar^{due north-1}

Product of the Geometric series

The Product of all the numbers nowadays in the geometric progression gives united states the overall production. Information technology is very useful while calculating the Geometric mean of the unabridged series.

Geometric Mean

By definition, it is the n^th root of Product of n numbers where 'n' denotes the number of terms present in the series. Geometric Mean differs from the Arithmetic Hateful every bit the latter is obtained by adding all terms and dividing by 'n', while the former is obtained past doing the product and and then taking the hateful of all the terms.

Significance of Geometric Mean

Geometric mean is calculated because it informs the compounding that is occurring from period to period. It tells the central beliefs of the Progression by taking the mean of Geometric progression. For example, The growth of bacteria can easily be analyzed using Geometric mean. In curt, Longer the Time Horizon or the values in the series differs from each other, the compounding becomes more critical, and hence, Geometric hateful is more appropriate to apply.

Formula for Geometric Hateful

$\bar{X}_{geom}=\sqrt [northward]{x_{1}.x_{2}.x_{iii}...x_{n}}$ ="277" manner="vertical-align: -8px;"/>

where,

$\bar{X}_{geom}=\text{symbol for Geometric mean}\\x_{1},x_{2},x_{3}...x_{n}=\text{terms present in the geometric series}\\n=\text{number of terms present in the series}$

Question ane: What is the Geometric hateful 2, 4, 8?

Answer:

According to the formula,

$=\sqrt [3]{(2)(4)(8)}\\=4$

Question 2: Notice the first term and mutual gene in the following Geometric Progression:

4, viii, 16, 32, 64,….

Answer:

Here,
is clear that the first term is 4, a=4
We obtain common Ratio past dividing 1st term from 2nd:

r = eight/4 = two

Question iii: Find the 8^thursday and the n^thursdayterm for the K.P: 3, 9, 27, 81,….

Answer:

Put n=8 for eight^th term in the formula: ar^{due north-i}

For the One thousand.P : 3, 9, 27, 81….

First term (a) = 3

Mutual Ratio (r) = 9/3 = 3

8^th term = 3(iii)^8-i = 3(iii)⁷ = 6561

North^thursday = 3(3)^northward-
up> = 3(iii)ⁿ(3)^-1
= 3ⁿ

Question 4: For the G.P. : ii, 8, 32,…. which term will give the value 131073?

Respond:

Assume that the value 131073 is the N^th term,

a = 2, r = 8/ii = 4

Northward^th term (a_northward) = 2(4)^n-1 = 131073

4^northward-1 = 131073/two = 65536

4^n-i = 65536 = 4⁸

Equating the Pw
since the base is aforementioned:
n-1 = 8

n = ix

Question 5: Discover the sum upwards to 5^thursday and N^thterm of the series: $1, \frac{1}{2},\frac{1}{4},\frac{1}{8}...$

Answer:

a= 1, r = ane/2

Sum of N terms for the Thou.P, ${S_n =\frac{a(1-r^n)}{1-r}}$

$= \frac{1(1-(\frac{1}{2})^n)}{1-\frac{1}{2}}$

$= 2 (1-(\frac{1}{2})^n)$

Sum of first 5 terms ⇒ a_v = $2 ( 1-(\frac{1}{2})^5)$

= $2 ( 1-(\frac{1}{32}))$

= $(\frac{31}{16})$ h="42" fashion="vertical-align: -9px;"/>

Question 6: Notice the Sum of the Space Chiliad.P: 0.v, i, ii, 4, 8, …

Respond:

Formula for the Sum of Infinite M.P: $\frac{a}{1-r}$ ; r≠0

a = 0.5, r = 2

Due south_∞= (0.5)/(i-2) = 0.5/(-one)= -0.v

Question 7: Find the sum of the Series: v, 55, 555, 5555,… n terms

Answer:

The given Series is non in G.P just it can easily be converted into a Yard.P with some elementary modifications.

Taking 5 mutual from the serial: five (ane, xi, 111, 1111,… n terms)

Dividing and Multiplying with 9: $\frac{5}{9}(9+ 99+ 999+...n terms)$

⇒ $\frac{5}{9}[((10+(10)^2+(10)^3+...n terms)-(1+1+1+...n terms)]$

⇒ $\frac{5}{9}[(\frac{10((10)^n-1)}{10-1})-(n)]$

⇒ $\frac{5}{9}[(\frac{10((10)^n-1)}{9})-(n)]$