
If ${S_n} = \dfrac{1}{{6.11}} + \dfrac{1}{{11.16}} + \dfrac{1}{{16.21}} + .......$ upto n then $6{S_n}$ is equals to
A. $\dfrac{{5n - 4}}{{5n + 6}}$
B. $\dfrac{n}{{5n + 6}}$
C. $\dfrac{{2n}}{{5n + 6}}$
D. $\dfrac{1}{{5n + 6}}$
Answer
217.8k+ views
Hint: We have to find the value of $6{S_n}$ using the value of ${S_n} = \dfrac{1}{{6.11}} + \dfrac{1}{{11.16}} + \dfrac{1}{{16.21}} + .......$. Firstly, we will simplify each term of ${S_n}$ to get a simplified pattern so that we can find the simplest value of ${S_n}$. Then we use that value to find $6{S_n}$. We simply multiply ${S_n}$ by $6$ to get the required answer.
Complete step by step Solution:
We have to find the value of $6{S_n}$
First we will simplify each term ${S_n} = \dfrac{1}{{6.11}} + \dfrac{1}{{11.16}} + \dfrac{1}{{16.21}} + .......$
$\dfrac{1}{{6.11}} = \dfrac{1}{5}\left( {\dfrac{1}{6} - \dfrac{1}{{11}}} \right)$
$\dfrac{1}{{11.16}} = \dfrac{1}{5}\left( {\dfrac{1}{{11}} - \dfrac{1}{{16}}} \right)$
$\dfrac{1}{{16.21}} = \dfrac{1}{5}\left( {\dfrac{1}{{16}} - \dfrac{1}{{21}}} \right)$
Similarly nth term
$\dfrac{1}{5}\left( {\dfrac{1}{{5n + 1}} - \dfrac{1}{{5n + 6}}} \right)$
${S_n} = \dfrac{1}{5}\left( {\dfrac{1}{6} - \dfrac{1}{{11}}} \right) + \dfrac{1}{5}\left( {\dfrac{1}{{11}} - \dfrac{1}{{16}}} \right) + \dfrac{1}{5}\left( {\dfrac{1}{{16}} - \dfrac{1}{{21}}} \right) + .......\dfrac{1}{5}\left( {\dfrac{1}{{5n + 1}} - \dfrac{1}{{5n + 6}}} \right)$
Taking $\dfrac{1}{5}$ common from previous
${S_n} = \dfrac{1}{5}\left( {\dfrac{1}{6} - \dfrac{1}{{11}} + \dfrac{1}{{11}} - \dfrac{1}{{16}} + \dfrac{1}{{16}} - \dfrac{1}{{21}} + .......\dfrac{1}{{5n + 1}} - \dfrac{1}{{5n + 6}}} \right)$
After simplifying, we get
${S_n} = \dfrac{1}{5}\left( {\dfrac{1}{6} - \dfrac{1}{{5n + 6}}} \right)$
${S_n} = \dfrac{1}{5}\left( {\dfrac{{5n + 6 - 6}}{{6(5n + 6)}}} \right)$
After solving,
${S_n} = \dfrac{1}{5}\left( {\dfrac{{5n}}{{6(5n + 6)}}} \right)$
${S_n} = \dfrac{n}{{6(5n + 6)}}$
Now, we multiply ${S_n}$ by 6 to get the final answer
$6{S_n} = \dfrac{{6n}}{{6(5n + 6)}}$
$6{S_n} = \dfrac{n}{{5n + 6}}$
Hence, the correct option is (d).
Additional Information:
In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity. . The study of series may be a major part of calculus and its generalization, mathematical analysis. Series area units are utilized in most areas of arithmetic, even for learning finite structures through generating functions. In additionally to their omnipresence in arithmetic, infinite series also are widely utilized in different quantitative disciplines like physics, engineering, statistics, and finance
Note: Students can make mistakes while calculating the value of ${S_n}$. They should pay attention while simplifying each term of ${S_n}$ to avoid any calculation error. Then after simplifying they should do further calculations carefully in order to get the correct answer. After that, they have carefully simplified the series ${S_n}$ so that they can get a simplified equation for further process. At last, they just have to multiply ${S_n}$ by 6.
Complete step by step Solution:
We have to find the value of $6{S_n}$
First we will simplify each term ${S_n} = \dfrac{1}{{6.11}} + \dfrac{1}{{11.16}} + \dfrac{1}{{16.21}} + .......$
$\dfrac{1}{{6.11}} = \dfrac{1}{5}\left( {\dfrac{1}{6} - \dfrac{1}{{11}}} \right)$
$\dfrac{1}{{11.16}} = \dfrac{1}{5}\left( {\dfrac{1}{{11}} - \dfrac{1}{{16}}} \right)$
$\dfrac{1}{{16.21}} = \dfrac{1}{5}\left( {\dfrac{1}{{16}} - \dfrac{1}{{21}}} \right)$
Similarly nth term
$\dfrac{1}{5}\left( {\dfrac{1}{{5n + 1}} - \dfrac{1}{{5n + 6}}} \right)$
${S_n} = \dfrac{1}{5}\left( {\dfrac{1}{6} - \dfrac{1}{{11}}} \right) + \dfrac{1}{5}\left( {\dfrac{1}{{11}} - \dfrac{1}{{16}}} \right) + \dfrac{1}{5}\left( {\dfrac{1}{{16}} - \dfrac{1}{{21}}} \right) + .......\dfrac{1}{5}\left( {\dfrac{1}{{5n + 1}} - \dfrac{1}{{5n + 6}}} \right)$
Taking $\dfrac{1}{5}$ common from previous
${S_n} = \dfrac{1}{5}\left( {\dfrac{1}{6} - \dfrac{1}{{11}} + \dfrac{1}{{11}} - \dfrac{1}{{16}} + \dfrac{1}{{16}} - \dfrac{1}{{21}} + .......\dfrac{1}{{5n + 1}} - \dfrac{1}{{5n + 6}}} \right)$
After simplifying, we get
${S_n} = \dfrac{1}{5}\left( {\dfrac{1}{6} - \dfrac{1}{{5n + 6}}} \right)$
${S_n} = \dfrac{1}{5}\left( {\dfrac{{5n + 6 - 6}}{{6(5n + 6)}}} \right)$
After solving,
${S_n} = \dfrac{1}{5}\left( {\dfrac{{5n}}{{6(5n + 6)}}} \right)$
${S_n} = \dfrac{n}{{6(5n + 6)}}$
Now, we multiply ${S_n}$ by 6 to get the final answer
$6{S_n} = \dfrac{{6n}}{{6(5n + 6)}}$
$6{S_n} = \dfrac{n}{{5n + 6}}$
Hence, the correct option is (d).
Additional Information:
In mathematics, a series is, roughly speaking, a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity. . The study of series may be a major part of calculus and its generalization, mathematical analysis. Series area units are utilized in most areas of arithmetic, even for learning finite structures through generating functions. In additionally to their omnipresence in arithmetic, infinite series also are widely utilized in different quantitative disciplines like physics, engineering, statistics, and finance
Note: Students can make mistakes while calculating the value of ${S_n}$. They should pay attention while simplifying each term of ${S_n}$ to avoid any calculation error. Then after simplifying they should do further calculations carefully in order to get the correct answer. After that, they have carefully simplified the series ${S_n}$ so that they can get a simplified equation for further process. At last, they just have to multiply ${S_n}$ by 6.
Recently Updated Pages
Arithmetic, Geometric & Harmonic Progressions Explained

Cartesian Form of Vector Explained: Formula, Examples & Uses

Apparent Frequency Explained: Formula, Uses & Examples

Calorimetry: Definition, Principles & Calculations

Centrifugal Force Explained: Definition, Formula & Examples

Charge in a Magnetic Field: Definition, Formula & Examples

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

How to Convert a Galvanometer into an Ammeter or Voltmeter

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

