
Which one is an example of A.P. property?
A) Constant $a$ is added to each term of an A.P. will form a new A.P. with different common difference.
B) Constant $a$ is subtracted to each term of an A. P. will form a new A.P. with different common differences.
C) Constant $a$ is divided to each term of an A.P. will not form a new A.P. with same common difference.
D) Constant $a$ is added to each term of an A.P. will form a new A.P. with the same common difference.
Answer
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Hint: We know that an AP or arithmetic progression is a sequence of numbers such that the difference of any two successive members is constant.
Note: We must consider the cases when a constant is added, subtracted and divided to each term. We must note that the constant is subtracted from each term and consider the order as subtraction is not commutative. We must understand that when a constant is added, subtracted, multiplied or divided to each term of an arithmetic progression will give a new AP.
$A,\,A + d,\,A + 2d,....$ be an AP with common difference $d$.
We can take a general arithmetic progression. We perform the required operations. Then we can compare with the given options and choose the correct answers.
Complete step by step solution:
The general sequence of Arithmetic progression is $A,\,A + d,\,A + 2d,....$ be an AP with common difference $d$.
Complete step by step solution:
The general sequence of Arithmetic progression is $A,\,A + d,\,A + 2d,....$ be an AP with common difference $d$.
Let's check options step by step.
A) Now we can add the constant as to each term. Then the AP will become,
$ \Rightarrow A + a,\,A + a + d,\,A + a + 2d,....$
Then the new common difference is given by, $A + a + 2d - \,\left( {A + a + d} \right) = d$
Therefore, constant a is added to each term of an A.P. will form a new A.P. with the same common difference.
$ \Rightarrow A + a,\,A + a + d,\,A + a + 2d,....$
Then the new common difference is given by, $A + a + 2d - \,\left( {A + a + d} \right) = d$
Therefore, constant a is added to each term of an A.P. will form a new A.P. with the same common difference.
This can be checked with an example.
Example:
Original A.P.: $5,10,15,20,\cdot \cdot \cdot $ with common difference $d=5$ and
Let's constant: $a=2$
Adding constant term to the A.P.
New A.P. will be $5+2,\, 10+2, \, 15+2, \, 20+2, \, ......$
New A.P. $7, 12, 17, 22....$, This clear that these terms are in A.P. with the same common difference (d=5) as the original A.P.
$\therefore$ The statement in option (A) "Constant $a$ is added to each term of an A.P. will form a new A.P. with different common difference" is not correct.
B) Now we can subtract the constant a from each term. Then the AP will become,
$ \Rightarrow A - a,\,A - a + d,\,A - a + 2d,....$
Then the new common difference is given by, $A - a + 2d - \,\left( {A - a + d} \right) = d$
$ \Rightarrow A - a,\,A - a + d,\,A - a + 2d,....$
Then the new common difference is given by, $A - a + 2d - \,\left( {A - a + d} \right) = d$
Example: Original A.P.: $5,10,15,20,\cdot \cdot \cdot $ with common difference $d=5$ and
Let's constant: $a=2$
Subtracting constant term from the A.P.,
New A.P. is $5-2, \, 10-2, \, 15-2. \, 20-2 ... $
New A.P. is $3, \, 8, \, 13, \, 18, \,..... $ It is also in A.P with the same common difference $d=5$ as original A.P.
New A.P. is $3, \, 8, \, 13, \, 18, \,..... $ It is also in A.P with the same common difference $d=5$ as original A.P.
$\therefore $ The statement in option (B) "Constant $a$ is subtracted to each term of an A. P. will form a new A.P. with different common differences." is incorrect.
C) Now we can divide each term of the AP with the constant a.
$ \Rightarrow \dfrac{A}{a},\,\dfrac{{A + d}}{a},\,\dfrac{{A + 2d}}{a},....$
Then the new common difference is given by,
$ \Rightarrow \dfrac{{A + 2d}}{a} - \dfrac{{A + d}}{a} = \dfrac{d}{a}$
Therefore, when constant a is divided to each term of an A.P. will not form a new A.P. with the same common difference.
By comparing the options, we can say that options C and D are true.
Example: Original A.P.: $5,10,15,20,\cdot \cdot \cdot $ with common difference $d=5$ and
$ \Rightarrow \dfrac{A}{a},\,\dfrac{{A + d}}{a},\,\dfrac{{A + 2d}}{a},....$
Then the new common difference is given by,
$ \Rightarrow \dfrac{{A + 2d}}{a} - \dfrac{{A + d}}{a} = \dfrac{d}{a}$
Therefore, when constant a is divided to each term of an A.P. will not form a new A.P. with the same common difference.
By comparing the options, we can say that options C and D are true.
Example: Original A.P.: $5,10,15,20,\cdot \cdot \cdot $ with common difference $d=5$ and
Let's constant: $a=5$
Dividing the terms with a constant term,New A.P. = $\dfrac{5}{5}, \, \dfrac{10}{5}, \, \dfrac{15}{5}, \, \dfrac{20}{5}, \,......$
New A.P. = $1, \, 2, \, 3, \, 4, \, ....$ common difference is $1$.
The new A.P. formed has the terms in A.P. But the common difference is different from the original A.P.
$\therefore$ The statement given in option (C), "Constant a is divided to each term of an A.P. will not form a new A.P. with same common difference." which is correct.
$\therefore$ The statement given in option (C), "Constant a is divided to each term of an A.P. will not form a new A.P. with same common difference." which is correct.
D) Constant $a$ is added to each term of an A.P. will form a new A.P. with the same common difference. This statement is True and is verified above in the verification of option (A).
Therefore, Both option (C) and (D) are correct.
Note: We must consider the cases when a constant is added, subtracted and divided to each term. We must note that the constant is subtracted from each term and consider the order as subtraction is not commutative. We must understand that when a constant is added, subtracted, multiplied or divided to each term of an arithmetic progression will give a new AP.
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