Answer
Verified
420.3k+ views
Hint:- In this question, given series is in Arithmetic –Geometric Progression(AGP) so we first convert it into Geometric Progression(GP) and then apply infinite series summation of GP($Su{m_\infty } = \dfrac{a}{{1 - r}}$) to get answer.
Complete step-by-step solution -
Arithmetic –Geometric Progression(AGP) is a progression in which each term can be represented as the product of the terms of an Arithmetic Progression(AP) and a Geometric Progression(GP).
Let $S = \dfrac{1}{{10}} + \dfrac{2}{{{{10}^2}}} + \dfrac{3}{{{{10}^3}}} + ..................$ eq.1
Since, in S numerator varies in AP while denominator varies in GP. Hence, it is in AGP.
Now, divide eq.1 with 10, we get
$ \Rightarrow \dfrac{S}{{10}} = \dfrac{1}{{{{10}^2}}} + \dfrac{2}{{{{10}^3}}} + \dfrac{3}{{{{10}^4}}}..................{\text{ eq}}{\text{.2}}$
Subtract eq.2 from eq.1, we get
$ \Rightarrow \dfrac{{9S}}{{10}} = \dfrac{1}{{{{10}^1}}} + \dfrac{1}{{{{10}^2}}} + \dfrac{1}{{{{10}^3}}} + \dfrac{1}{{{{10}^4}}}..................{\text{ eq}}{\text{.3}}$
Multiple eq.3 with 10, we get
$ \Rightarrow 9S = 1 + \dfrac{1}{{{{10}^1}}} + \dfrac{1}{{{{10}^2}}} + \dfrac{1}{{{{10}^3}}} + \dfrac{1}{{{{10}^4}}}..................{\text{ eq}}{\text{.4}}$
We can observe that RHS of eq.4 is in Geometric Progression(GP). We know that sum of infinite GP series is given by
$ \Rightarrow Su{m_\infty } = \dfrac{a}{{1 - r}}$
Where $
a = {\text{first term of GP}} \\
r = {\text{ common ratio}} \\
$
From eq.4 we can observe that $a = 1{\text{ and }}r = \dfrac{1}{{10}}$
Then, $
\Rightarrow 9{S_\infty } = \dfrac{1}{{1 - \dfrac{1}{{10}}}} \\
\Rightarrow 9{S_\infty } = {\text{ }}\dfrac{{10}}{9} \\
\Rightarrow {S_\infty } = {\text{ }}\dfrac{{10}}{{81}} \\
$
Hence, option B. is correct.
Note:- Whenever we get this type of question first we need to observe that the given series is in AGP or AP or GP. If it is in AGP we convert it into GP by performing simple operations. Later we have to apply the concept of AGP and infinite series summation formula of GP(${S_\infty } = \dfrac{a}{{1 - r}}$).
Complete step-by-step solution -
Arithmetic –Geometric Progression(AGP) is a progression in which each term can be represented as the product of the terms of an Arithmetic Progression(AP) and a Geometric Progression(GP).
Let $S = \dfrac{1}{{10}} + \dfrac{2}{{{{10}^2}}} + \dfrac{3}{{{{10}^3}}} + ..................$ eq.1
Since, in S numerator varies in AP while denominator varies in GP. Hence, it is in AGP.
Now, divide eq.1 with 10, we get
$ \Rightarrow \dfrac{S}{{10}} = \dfrac{1}{{{{10}^2}}} + \dfrac{2}{{{{10}^3}}} + \dfrac{3}{{{{10}^4}}}..................{\text{ eq}}{\text{.2}}$
Subtract eq.2 from eq.1, we get
$ \Rightarrow \dfrac{{9S}}{{10}} = \dfrac{1}{{{{10}^1}}} + \dfrac{1}{{{{10}^2}}} + \dfrac{1}{{{{10}^3}}} + \dfrac{1}{{{{10}^4}}}..................{\text{ eq}}{\text{.3}}$
Multiple eq.3 with 10, we get
$ \Rightarrow 9S = 1 + \dfrac{1}{{{{10}^1}}} + \dfrac{1}{{{{10}^2}}} + \dfrac{1}{{{{10}^3}}} + \dfrac{1}{{{{10}^4}}}..................{\text{ eq}}{\text{.4}}$
We can observe that RHS of eq.4 is in Geometric Progression(GP). We know that sum of infinite GP series is given by
$ \Rightarrow Su{m_\infty } = \dfrac{a}{{1 - r}}$
Where $
a = {\text{first term of GP}} \\
r = {\text{ common ratio}} \\
$
From eq.4 we can observe that $a = 1{\text{ and }}r = \dfrac{1}{{10}}$
Then, $
\Rightarrow 9{S_\infty } = \dfrac{1}{{1 - \dfrac{1}{{10}}}} \\
\Rightarrow 9{S_\infty } = {\text{ }}\dfrac{{10}}{9} \\
\Rightarrow {S_\infty } = {\text{ }}\dfrac{{10}}{{81}} \\
$
Hence, option B. is correct.
Note:- Whenever we get this type of question first we need to observe that the given series is in AGP or AP or GP. If it is in AGP we convert it into GP by performing simple operations. Later we have to apply the concept of AGP and infinite series summation formula of GP(${S_\infty } = \dfrac{a}{{1 - r}}$).
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
What is the type of food and mode of feeding of the class 11 biology CBSE
Name 10 Living and Non living things class 9 biology CBSE