
Find the value of $p$ for which the numbers $2p-1,3p+1,11$ are in A.P. and then find the arithmetic progression.
Answer
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Hint: For solving this question we will directly use the basic concept of A.P. and that is the difference between the consecutive terms should be equal. After that, we will see the condition for any three numbers $a,b,c$ to be in A.P. and we will apply the result here to find the suitable value of $p$ . Moreover, we will also verify our answer.
Complete step-by-step answer:
It is given that three numbers $2p-1,3p+1,11$ are in A.P. and we have to find a suitable value of $p$ . After that, we have to write that arithmetic progression.
Now, before we proceed we should know if any three numbers $a,b,c$ are in A.P. then what is the sufficient condition for that. As we know that in A.P. the difference between consecutive terms is equal. Then,
$\begin{align}
& b-a=c-b \\
& \Rightarrow 2b=a+c.................\left( 1 \right) \\
\end{align}$
Now, from the above result, we conclude that if three numbers $a,b,c$ then $2b=a+c$ . And it is given that three numbers $2p-1,3p+1,11$ are in A.P. so, we can put the value of $a=2p-1$ , $b=3p+1$ and $c=11$ in the equation (1). Then,
$\begin{align}
& 2b=a+c \\
& \Rightarrow 2\left( 3p+1 \right)=2p-1+11 \\
& \Rightarrow 6p+2-2p=10 \\
& \Rightarrow 4p=8 \\
& \Rightarrow p=2 \\
\end{align}$
Now, from the above result, we conclude that if the three numbers $2p-1,3p+1,11$ are in A.P. then the value of $p$ should be equal to 2. And the three numbers will be 3, 7, 11.
Thus, the given three numbers will be 3, 7 and 11 and we can verify it by checking the value of the difference between the consecutive terms whether it is equal or not.
Verification:
$\begin{align}
& 7-3=4 \\
& 11-7=4 \\
\end{align}$
Hence verified.
Note: Here, the student should first understand what is asked in the problem. After that, directly apply the basic concept of A.P. that is the difference of the consecutive terms should be equal. Moreover, though the problem is very easy, we should avoid calculation mistakes while solving to get the correct answer.
Complete step-by-step answer:
It is given that three numbers $2p-1,3p+1,11$ are in A.P. and we have to find a suitable value of $p$ . After that, we have to write that arithmetic progression.
Now, before we proceed we should know if any three numbers $a,b,c$ are in A.P. then what is the sufficient condition for that. As we know that in A.P. the difference between consecutive terms is equal. Then,
$\begin{align}
& b-a=c-b \\
& \Rightarrow 2b=a+c.................\left( 1 \right) \\
\end{align}$
Now, from the above result, we conclude that if three numbers $a,b,c$ then $2b=a+c$ . And it is given that three numbers $2p-1,3p+1,11$ are in A.P. so, we can put the value of $a=2p-1$ , $b=3p+1$ and $c=11$ in the equation (1). Then,
$\begin{align}
& 2b=a+c \\
& \Rightarrow 2\left( 3p+1 \right)=2p-1+11 \\
& \Rightarrow 6p+2-2p=10 \\
& \Rightarrow 4p=8 \\
& \Rightarrow p=2 \\
\end{align}$
Now, from the above result, we conclude that if the three numbers $2p-1,3p+1,11$ are in A.P. then the value of $p$ should be equal to 2. And the three numbers will be 3, 7, 11.
Thus, the given three numbers will be 3, 7 and 11 and we can verify it by checking the value of the difference between the consecutive terms whether it is equal or not.
Verification:
$\begin{align}
& 7-3=4 \\
& 11-7=4 \\
\end{align}$
Hence verified.
Note: Here, the student should first understand what is asked in the problem. After that, directly apply the basic concept of A.P. that is the difference of the consecutive terms should be equal. Moreover, though the problem is very easy, we should avoid calculation mistakes while solving to get the correct answer.
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