Question

If the numbers $\left( 2n-1 \right),\left( 3n-2 \right)$and $\left( 6n-2 \right)$ are in AP, find the value of $n$ and the numbers.

Answer 1

Hint: We use the fact that in an arithmetic progression AP any two consecutive terms will have the same difference. We take the difference of $\left( 2n-1 \right),\left( 3n-2 \right)$ and the difference of $\left( 3n+2 \right),\left( 6n-2 \right)$and then equate them. We find $n$ by solving for $n$ and then put the value of $n$ in $\left( 2n-1 \right),\left( 3n-2 \right)$and $\left( 6n-2 \right)$ to get the numbers.

Complete step-by-step answer:
A sequence is defined as the enumerated collection of numbers where repetitions are allowed and order of the numbers matters. The members of the sequence are called terms. Mathematically, a sequence with infinite terms is written as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...\]
If the sequence has finite terms terminated by a term then we write the sequence as
\[\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...{{x}_{n}}\]
Arithmetic sequence otherwise known as arithmetic progression, abbreviated as AP is a type sequence where the difference between any two consecutive terms is constant. If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an AP, then ${{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$ .
We are given in the question three consecutive terms of an AP as expressions in $n$ as $\left( 2n-1 \right),\left( 3n-2 \right)$and $\left( 6n-2 \right)$. Since the difference of two consecutive terms in AP are equal the difference between $\left( 2n-1 \right),\left( 3n-2 \right)$ and $\left( 3n+2 \right),\left( 6n-2 \right)$ will be equal. So we have;
\[\left( 3n+2 \right)-\left( 2n-1 \right)=\left( 6n-1 \right)-\left( 3n+2 \right)\]
We remove the bracket following BOMAS rule to have;
\[\begin{align}
  & \Rightarrow 3n+2-2n+1=6n-1-3n-2 \\
 & \Rightarrow n+3=3n-3 \\
\end{align}\]
We add 3 both side and then subtract $2n$ both side to have;
\[\Rightarrow 2n=6\]
We divide $n$both side of the equation to have;
\[\Rightarrow n=3\]
We put the value of $n=2$ and find the numbers given in the question
\[\begin{align}
  & 2n-1=2\times 3-1=5 \\
 & 3n+2=3\times 3+1=11 \\
 & 6n-1=6\times 3-1=17 \\
\end{align}\]
So the value of $n$ is 3 and the numbers in AP are $5,11,17$.

Note: The difference between two terms is called common difference and denoted as $d$ where$d={{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$. . Here in this problem the common difference is $d=11-5=17-11=6$. We must be careful of confusion between AP from geometric progression (GP) where there is a common ratio instead of common difference between any two terms.