
If \[a,b,c,d,e,f\] are in A.P., then find the value of \[e - c\].
A. \[2\left( {c - a} \right)\]
B. \[2\left( {f - d} \right)\]
C. \[2\left( {d - c} \right)\]
D. \[d - c\]
Answer
163.2k+ views
Hint: Apply the general term formula to find the value of \[a,b,c,d,e,f\]. Then find the common difference of each consecutive term. Then we will calculate the value of \[e - c\].
Formula used:
The general terms of an AP are \[a,a + d,a + 2d,a + 3d,a + 4d,a + 5d, \cdots \]
The common difference of an AP is \[{n^{th}}{\rm{term}} - {\left( {n - 1} \right)^{th}}{\rm{term}}\].
Complete step by step solution:
Given that \[a,b,c,d,e,f\] are in A.P.
Let the common difference of the AP be \[p\].
Apply the general term formula of an A.P.
The first term is \[a\].
The second term is \[b = a + p\]
The third term is \[c = a + 2p\]
The fourth term is \[d = a + 3p\]
The fifth term is \[e = a + 4p\]
The sixth term is \[f = a + 5p\]
We know that the common difference is the difference between two consecutive terms.
So, \[b - a = c - b = d - c = e - d = f - e = p\]
Calculate the value of \[e - c\]
\[e - c\]
Putting the value of \[e\] and \[c\]
\[ = a + 4p - \left( {a + 2p} \right)\]
Open parenthesis
\[ = a + 4p - a - 2p\]
Simplify the above expression
\[ = 2p\]
We know that \[b - a = c - b = d - c = e - d = f - e = p\]
So \[2p = 2\left( {b - a} \right) = 2\left( {c - b} \right) = 2\left( {d - c} \right) = 2\left( {e - d} \right) = 2\left( {f - e} \right)\]
Thus \[e - c = 2\left( {d - c} \right)\].
Hence option C is the correct option.
Note: Many students make mistakes to calculate the common difference. They calculate the common difference like \[a - b = b - c = c - d = d - e = e - f\] which is incorrect. The correct way is \[b - a = c - b = d - c = e - d = f - e = p\].
Formula used:
The general terms of an AP are \[a,a + d,a + 2d,a + 3d,a + 4d,a + 5d, \cdots \]
The common difference of an AP is \[{n^{th}}{\rm{term}} - {\left( {n - 1} \right)^{th}}{\rm{term}}\].
Complete step by step solution:
Given that \[a,b,c,d,e,f\] are in A.P.
Let the common difference of the AP be \[p\].
Apply the general term formula of an A.P.
The first term is \[a\].
The second term is \[b = a + p\]
The third term is \[c = a + 2p\]
The fourth term is \[d = a + 3p\]
The fifth term is \[e = a + 4p\]
The sixth term is \[f = a + 5p\]
We know that the common difference is the difference between two consecutive terms.
So, \[b - a = c - b = d - c = e - d = f - e = p\]
Calculate the value of \[e - c\]
\[e - c\]
Putting the value of \[e\] and \[c\]
\[ = a + 4p - \left( {a + 2p} \right)\]
Open parenthesis
\[ = a + 4p - a - 2p\]
Simplify the above expression
\[ = 2p\]
We know that \[b - a = c - b = d - c = e - d = f - e = p\]
So \[2p = 2\left( {b - a} \right) = 2\left( {c - b} \right) = 2\left( {d - c} \right) = 2\left( {e - d} \right) = 2\left( {f - e} \right)\]
Thus \[e - c = 2\left( {d - c} \right)\].
Hence option C is the correct option.
Note: Many students make mistakes to calculate the common difference. They calculate the common difference like \[a - b = b - c = c - d = d - e = e - f\] which is incorrect. The correct way is \[b - a = c - b = d - c = e - d = f - e = p\].
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Instantaneous Velocity - Formula based Examples for JEE

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

JEE Advanced 2025 Notes
