
The fourth term of an A.P. is three times of the first term and the seventh term exceeds the twice of the third term by one, then find the common difference of the progression.
A. $2$
B. $3$
C. $\dfrac{3}{2}$
D. $-1$
Answer
162.9k+ views
Hint: In this question, we are to find the common difference in the given arithmetic series. Here we have provided with some of the terms and their relation with one another. By comparing those terms with the general terms of the arithmetic series, we get the required values.
Formula Used: An arithmetic series is written as
$a+(a+d)+(a+2d)+...+(a+(n-1)d)+...$
Here the first term is represented by $a$ and the common difference is represented by $d$.
The $nth$ term of an arithmetic series is written as
${{t}_{n}}=a+(n-1)d$
The sum of the arithmetic series of $n$ terms is
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
The common difference is calculated by the formula $d={{a}_{n}}-{{a}_{n-1}}$
Complete step by step solution: Consider the terms of the series with arithmetic progression as
$a+(a+d)+(a+2d)+...+(a+(n-1)d)+...$
The given terms are:
The first term: $a$
The third term: $(a+2d)$
The fourth term: $(a+3d)$
The seventh term: $(a+6d)$
It is given that,
The fourth term of the series is three times the first term i.e.,
$\begin{align}
& (a+3d)=3a \\
& \Rightarrow 3d=3a-a=2a \\
& \Rightarrow 3d=2a\text{ }...(1) \\
\end{align}$
The seventh term exceeds the twice of third term by one i.e.,
$\begin{align}
& (a+6d)=2(a+2d)+1 \\
& \Rightarrow a+6d=2a+4d+1 \\
& \Rightarrow 6d-4d=2a-a+1 \\
& \Rightarrow 2d=a+1\text{ }...(2) \\
\end{align}$
Then, on solving (2) and (1), we get
\[\begin{align}
& 3d=2a\Rightarrow d=\dfrac{2}{3}a \\
& \Rightarrow 2\left( \dfrac{2}{3}a \right)=a+1 \\
& \Rightarrow \dfrac{4}{3}a-a=1 \\
& \Rightarrow \dfrac{a}{3}=1 \\
& \therefore a=3 \\
\end{align}\]
On substituting the obtained value in (1), we get
$\begin{align}
& 3d=2(3) \\
& \Rightarrow 3d=6 \\
& \therefore d=2 \\
\end{align}$
Option ‘A’ is correct
Note: In such type of question, we have to consider the given terms with their general form and by simplifying them, we get the required values.
Formula Used: An arithmetic series is written as
$a+(a+d)+(a+2d)+...+(a+(n-1)d)+...$
Here the first term is represented by $a$ and the common difference is represented by $d$.
The $nth$ term of an arithmetic series is written as
${{t}_{n}}=a+(n-1)d$
The sum of the arithmetic series of $n$ terms is
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$
The common difference is calculated by the formula $d={{a}_{n}}-{{a}_{n-1}}$
Complete step by step solution: Consider the terms of the series with arithmetic progression as
$a+(a+d)+(a+2d)+...+(a+(n-1)d)+...$
The given terms are:
The first term: $a$
The third term: $(a+2d)$
The fourth term: $(a+3d)$
The seventh term: $(a+6d)$
It is given that,
The fourth term of the series is three times the first term i.e.,
$\begin{align}
& (a+3d)=3a \\
& \Rightarrow 3d=3a-a=2a \\
& \Rightarrow 3d=2a\text{ }...(1) \\
\end{align}$
The seventh term exceeds the twice of third term by one i.e.,
$\begin{align}
& (a+6d)=2(a+2d)+1 \\
& \Rightarrow a+6d=2a+4d+1 \\
& \Rightarrow 6d-4d=2a-a+1 \\
& \Rightarrow 2d=a+1\text{ }...(2) \\
\end{align}$
Then, on solving (2) and (1), we get
\[\begin{align}
& 3d=2a\Rightarrow d=\dfrac{2}{3}a \\
& \Rightarrow 2\left( \dfrac{2}{3}a \right)=a+1 \\
& \Rightarrow \dfrac{4}{3}a-a=1 \\
& \Rightarrow \dfrac{a}{3}=1 \\
& \therefore a=3 \\
\end{align}\]
On substituting the obtained value in (1), we get
$\begin{align}
& 3d=2(3) \\
& \Rightarrow 3d=6 \\
& \therefore d=2 \\
\end{align}$
Option ‘A’ is correct
Note: In such type of question, we have to consider the given terms with their general form and by simplifying them, we get the required values.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NEET 2025 – Every New Update You Need to Know
