Which of the following is a correct statement?
(A) \[{{Q}_{1}},{{Q}_{2}},{{Q}_{3}},......\] are in A.P. with common difference 5
(B) \[{{Q}_{1}},{{Q}_{2}},{{Q}_{3}},......\] are in A.P. with common difference 6
(C) \[{{Q}_{1}},{{Q}_{2}},{{Q}_{3}},......\] are in A.P. with common difference 11
(d) \[{{Q}_{1}}={{Q}_{2}},={{Q}_{3}}=......\]
Answer
542.4k+ views
Hint: For solving this question we should know about the Arithmetic Progression (A.P.). The A.P. is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value. It means the difference between them is always constant and this is showing that all have common differences.
Complete step-by-step solution:
According to our question we have to find the right statement from the given statements.
If we consider any term \[{{T}_{r}}\] for calculating the common difference and this will be calculating by assuming just next term of \[{{T}_{r}}\] which will be \[{{T}_{r+1}}\] and then we subtract to get the \[{{Q}_{r}}\] and then we again find \[{{Q}_{r+1}}\] and is we subtract to both then the common difference will be determined.
So, if \[{{T}_{r}}=3{{r}^{2}}+2r-1\]
And \[{{T}_{r+1}}=3{{\left( r+1 \right)}^{2}}+2\left( r+1 \right)-1\]
\[\begin{align}
& =3\left( {{r}^{2}}+2r+1 \right)+2r+2-1 \\
& =3{{r}^{2}}+6r+3+2r+2-1 \\
& =3{{r}^{2}}+8r+4 \\
\end{align}\]
So, the \[{{Q}_{r}}={{T}_{r+1}}-{{T}_{r}}\]
\[\begin{align}
& =\left( 3{{r}^{2}}+3r+4 \right)-\left( 3{{r}^{2}}+2r-1 \right) \\
& =3{{r}^{2}}+8r+4-3{{r}^{2}}-2r+1 \\
& =6r+5 \\
\end{align}\]
And if we find \[{{Q}_{r+1}}\]
Then, \[{{Q}_{r+1}}=6\left( r+1 \right)+5\]
\[\begin{align}
& =6r+6+5 \\
& =6r+11 \\
\end{align}\]
Thus, the common difference between both \[{{Q}_{r}}\] & \[{{Q}_{r+1}}\] is:
\[{{Q}_{r+1}}-{{Q}_{r}}=\left( 6r+11 \right)-\left( 6r+5 \right)=6\]
So, \[{{Q}_{1}},{{Q}_{2}},{{Q}_{3}},......\] are in A.P. with common difference 6. And (B) option is correct.
Note: The Arithmetic Progression properties are useful in these questions. Any term of Arithmetic Progression beside the first term, all the terms are exactly half of the addition of both values of right and left side of the given value. And so that they have common differences.
Complete step-by-step solution:
According to our question we have to find the right statement from the given statements.
If we consider any term \[{{T}_{r}}\] for calculating the common difference and this will be calculating by assuming just next term of \[{{T}_{r}}\] which will be \[{{T}_{r+1}}\] and then we subtract to get the \[{{Q}_{r}}\] and then we again find \[{{Q}_{r+1}}\] and is we subtract to both then the common difference will be determined.
So, if \[{{T}_{r}}=3{{r}^{2}}+2r-1\]
And \[{{T}_{r+1}}=3{{\left( r+1 \right)}^{2}}+2\left( r+1 \right)-1\]
\[\begin{align}
& =3\left( {{r}^{2}}+2r+1 \right)+2r+2-1 \\
& =3{{r}^{2}}+6r+3+2r+2-1 \\
& =3{{r}^{2}}+8r+4 \\
\end{align}\]
So, the \[{{Q}_{r}}={{T}_{r+1}}-{{T}_{r}}\]
\[\begin{align}
& =\left( 3{{r}^{2}}+3r+4 \right)-\left( 3{{r}^{2}}+2r-1 \right) \\
& =3{{r}^{2}}+8r+4-3{{r}^{2}}-2r+1 \\
& =6r+5 \\
\end{align}\]
And if we find \[{{Q}_{r+1}}\]
Then, \[{{Q}_{r+1}}=6\left( r+1 \right)+5\]
\[\begin{align}
& =6r+6+5 \\
& =6r+11 \\
\end{align}\]
Thus, the common difference between both \[{{Q}_{r}}\] & \[{{Q}_{r+1}}\] is:
\[{{Q}_{r+1}}-{{Q}_{r}}=\left( 6r+11 \right)-\left( 6r+5 \right)=6\]
So, \[{{Q}_{1}},{{Q}_{2}},{{Q}_{3}},......\] are in A.P. with common difference 6. And (B) option is correct.
Note: The Arithmetic Progression properties are useful in these questions. Any term of Arithmetic Progression beside the first term, all the terms are exactly half of the addition of both values of right and left side of the given value. And so that they have common differences.
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